\(\frac{dy}{dx}=3x^2-5x\). At \(x=1.5\), the gradient is \(3(1.5)^2-5(1.5)=6.75-7.5=-0.75\).

(a) \((x-3)^2=x^2-6x+9\), so \((x+5)(x^2-6x+9)=x^3-x^2-21x+45\).

(b) The graph crosses the \(x\)-axis at \((-5,0)\), touches it at the repeated root \((3,0)\), and crosses the \(y\)-axis at \((0,45)\). Its positive leading coefficient means it falls to the left and rises to the right.

(a) \(\frac{dy}{dx}=3x^2-12x=3x(x-4)\). At a turning point, \(dy/dx=0\), so \(x=0\) or \(x=4\). The y-coordinates are \(7\) and \(64-96+7=-25\). Thus the turning points are \(\boxed{(0,7)\text{ and }(4,-25)}\).

(b) The derivative is positive for \(x<0\), negative for \(0<x<4\), and positive for \(x>4\). The curve therefore rises, falls, then rises: \(\boxed{(0,7)\text{ is a local maximum}}\) and \(\boxed{(4,-25)\text{ is a local minimum}}\).

(a) \(dy/dx=4x-11\), so the tangent gradient at 2 is \(8-11=\boxed{-3}\).

(b) The perpendicular gradient is \(1/3\). At \(x=2\), \(y=8-22+12=-2\). Thus \(y+2=\frac13(x-2)\), giving \(\boxed{y=\frac13x-\frac83}\).

(a) At \(x=-1\), \(y=-1-2-2=\boxed{-5}\). At \(x=1\), \(y=1-2-2=\boxed{-3}\). At \(x=2.5\), \(y=15.625-12.5-2=1.125\), giving \(\boxed{1.1}\) to 1 decimal place.

(b) Plot all nine pairs and draw a smooth cubic curve through them.

(c) Rearranging the equation gives \(x^3-2x^2-2=x-3\). Draw \(\boxed{y=x-3}\) on the same axes and read the three intersection x-coordinates: approximately \(\boxed{-0.80,\ 0.55,\ 2.25}\).

(a) At \(x=-2\), \(y=3/4-5=-4.25\), giving \(\boxed{-4.3}\) to 1 decimal place. At \(x=-1\), \(y=3/2-5=\boxed{-3.5}\). At \(x=0\), \(y=3-5=\boxed{-2}\).

(b) Plot the table points and join them with a smooth increasing exponential curve.

(c) \(3\times2^x=7\) corresponds to \(y=7-5=2\). Draw \(\boxed{y=2}\) and read the intersection, \(\boxed{x\approx1.22}\).

(d) As x decreases, \(2^x\) approaches 0, so the horizontal asymptote is \(\boxed{y=-5}\).

(a) \(dy/dx=24x^2-6\). At a turning point, \(24x^2-6=0\), so \(x^2=1/4\) and \(x=\pm1/2\). Substitution into y gives \(\boxed{(-1/2,7)}\) and \(\boxed{(1/2,3)}\).

(b) \(d^2y/dx^2=48x\). At \(x=-1/2\), it is \(-24<0\), so \((-1/2,7)\) is a maximum. At \(x=1/2\), it is \(24>0\), so \((1/2,3)\) is a minimum.

(a) \(dy/dx=3x^2+2x-1=(3x-1)(x+1)\). Set this to zero: \(x=1/3\) or \(-1\). At \(x=-1\), \(y=-1+1+1=1\), giving \(\boxed{(-1,1)}\).

(b) The positive cubic rises to \((-1,1)\), falls to \((1/3,-5/27)\), then rises again. Thus the given point is a minimum.

(c) A horizontal line \(y=k\) meets the curve three times only when \(-5/27<k<1\). At either boundary there are two distinct intersections, and beyond them one. Therefore \(\boxed{k\le-5/27\text{ or }k\ge1}\).

(a) Draw the straight line from P that just touches the curve. It touches at \((1,2)\), so \(\boxed{a=1,b=2}\).

(b) Its gradient is \(m=(2-11)/(1-(-2))=-9/3=-3\). Through \((1,2)\): \(y-2=-3(x-1)\), giving \(\boxed{y=-3x+5}\).

(a) Differentiating gives \(nx^{n-1}+2qx+9\). Match terms: \(n-1=2\), so \(\boxed{n=3}\), and \(2q=-12\), so \(\boxed{q=-6}\).

(b) Set \(3x^2-12x+9=3(x-1)(x-3)=0\). Hence \(x=1\) or 3. Substitute in \(y=x^3-6x^2+9x\): \(y(1)=4,y(3)=0\). The turning points are \(\boxed{(1,4)\text{ and }(3,0)}\).

(a) Plot \((0,1),(1,3),(2,5),(3,7)\) and join them with a straight line.

(b) The lines intersect at \(\boxed{x=2,\ y=5}\). Check: \(2+5=7\) and \(5=2(2)+1\).

(a) \(2/x-1=0\), so \(\boxed{(2,0)}\).

(b) \(\boxed{x=0\text{ and }y=-1}\).

(c) Rearrange as \(2/x-1=x\) and draw \(y=x\). The intersections give \(\boxed{x=-2\text{ or }1}\). Check by multiplying by \(x\ne0\): \(x^2+x-2=(x+2)(x-1)=0\).

(a) \(3x-x^3=x(3-x^2)=0\). The nonzero roots are \(\boxed{x_A=-\sqrt3,x_B=\sqrt3}\).

(b)(i) \(\boxed{dy/dx=3-3x^2}\).

(b)(ii) Set \(3-3x^2=0\), giving \(x=\pm1\). At \(-1\), \(y=-3+1=-2\); at 1, \(y=3-1=2\). Hence \(\boxed{P=(-1,-2),Q=(1,2)}\).

(a) \(\boxed{dy/dx=3x^2-6x}\).

(b) \(3x(x-2)=0\), so \(x=0\) or 2. Then \(y(0)=1,y(2)=8-12+1=-3\). Thus \(\boxed{(0,1)\text{ and }(2,-3)}\).

Differentiate: \(dy/dx=11ax^{10}+3bx^{b-1}\). Compare coefficients and powers: \(11a=44\), \(3b=18\), \(c=b-1\). Thus \(\boxed{a=4,b=6,c=5}\).

(a) The missing values at \(x=-1,1,4\) are \(\boxed{2,0,12}\).

(b) Plot all eight points and join with a smooth cubic curve.

(c) Rearrange the equation as \(5x^2-x^3-4=10-x\). Draw \(y=10-x\); its three intersections with the cubic give \(\boxed{x\approx-1.5,\ 2,\ 4.5}\).

(a) At x=−1.5, y=2.625, giving \(\boxed{2.63}\); at x=1.5, y=3.375, giving \(\boxed{3.38}\). (b) Plot the nine points and join with a smooth curve.

(c) Rearrange as \(x^3-2x+3=2+0.5x\). Draw \(y=2+0.5x\), for example through (−2,1) and (2,3). The three intersections give approximately \(\boxed{x=-1.75,\ 0.43,\ 1.32}\).

(a) Differentiate each term: \(\boxed{dy/dx=3x^2+6x-13}\). (b) At x=3 the gradient is \(3(9)+6(3)-13=\boxed{32}\).

(a) Missing y values are \(\boxed{-2.5,-1.25,5.5}\). (b) Plot all nine points and join with a smooth cubic curve. (c) Rearrange as \(2x^3+6x^2-2.5=2\). Draw y=2; its intersections give approximately \(\boxed{x=-2.69,-1.08,0.77}\).

(d) A horizontal line has two distinct intersections when it passes through either turning point. They are at (−2,5.5) and (0,−2.5), so \(\boxed{k=-2.5\text{ or }5.5}\).

(a) Draw a straight line through (0,4) and (4/3,0), with negative gradient. (b) Draw a downward parabola, symmetric about the y-axis with vertex (0,0).

(c)(i) \(dy/dx=18x-6x^2=6x(3-x)\). Set it to zero: x=0 or x=3. Substitution gives \(\boxed{(0,10)\text{ and }(3,37)}\). (ii) \(d^2y/dx^2=18-12x\). At x=0 it is 18>0, so (0,10) is a minimum; at x=3 it is −18<0, so (3,37) is a maximum.

Draw \(y=2x\), for example through (0,0) and (1,2). Read the two intersections in the given interval: approximately \(\boxed{x=-0.52\text{ or }0.88}\).

(a) Draw a straight line just touching the curve near (3,5.4), following its local direction. (b) For example, read approximately (2.5,2.7) and (4,10.8) on the tangent. Gradient≈\((10.8-2.7)/(4-2.5)=\boxed{5.4}\). A graphical estimate varies slightly.

(a) \(x^2+7x-18=(x+9)(x-2)\). Upward parabola through \(\boxed{(-9,0),(2,0),(0,-18)}\), with minimum(−3.5,−30.25). (b)(i) \(\boxed{dy/dx=2x-3}\). (ii) Set 2x−3=0; x=1.5 and y=−30.25: \(\boxed{(1.5,-30.25)}\). (c) Equate: \(x^2-3x-28=5-2x\), so \(x^2-x-33=0\). Formula gives \(x=(1\pm\sqrt{133})/2\). Substitute unrounded x into 5−2x to obtain \(\boxed{(-5.27,15.53),(6.27,-7.53)}\).

(a) At x=−1, \(-4(-1)^3-9(-1)^2+5=4-9+5=0\). The gradient is zero, so the point is stationary. (b)(i) \(dy/dx=6x^2-6\). At x=−2 this is 24−6=\(\boxed{18}\). (ii) Set \(6x^2-6=0\), so \(\boxed{x=1\text{ or }-1}\).

(a) \(\boxed{dy/dx=7x^6-42x^5}\). (b) At x=−1,y=−8 and gradient=7+42=49. \(y+8=49(x+1)\), so \(\boxed{y=49x+41}\). (c) \(7x^5(x-6)=0\), giving x=0 or 6. Substitution gives \(\boxed{(0,0),(6,-46656)}\). The derivative changes from positive to negative at 0 and negative to positive at 6, so these are a maximum and minimum.

(a)(i) \(\boxed3\). (ii) Read the intersections with y=5: \(\boxed{x\approx-2,-0.41,2.41}\). (iii) The left branch just touches y=3 and the right branch crosses it, so \(\boxed{k=3}\). (iv) \(\boxed{\text{asymptote}}\). (b)(i) Draw the line through(0,−2),(2,0). (ii) It meets the curve at \(\boxed{x=1}\). (c) Using the exact graph point(2,3): \(3=4-c/2\), so \(\boxed{c=2}\). (d) \(x^2-2/x=x-2\). Multiply by x and rearrange: \(x^3-x^2+2x=2\), so \(\boxed{p=-1,q=2}\).

(a) \((x+2)(x^2-2x+1)=x^3-2x^2+x+2x^2-4x+2=x^3-3x+2\). (b) On the curve, \(x^3=y+3x-2\). Hence \(2(y+3x-2)-5x=0\), or \(y=2-\frac12x\). Draw this line and read the intersections: \(\boxed{x\approx-1.58,\ 0,\ 1.58}\).

(a) \(\boxed{dy/dx=3x^2-18x-48}\). (b) Set derivative zero: \(3(x-8)(x+2)=0\), so x=−2 or 8. The points are \(\boxed{(-2,52),(8,-448)}\). (c) \(d^2y/dx^2=6x-18\). At x=−2 it is −30<0, so (−2,52) is a maximum; at x=8 it is30>0, so (8,−448) is a minimum.

(a) At x=−4, y=−64+64−4=\(\boxed{-4}\); at x=−1, y=−1+4−4=\(\boxed{-1}\). (b) Plot all eight points and join with a smooth cubic curve. (c) The tangent shown passes through (1,1) and (0,−10). Gradient=(1−(−10))/(1−0)=\(\boxed{11}\). (d) Since the curve has y=x³+4x²−4, the equation is y−x−2=0. Draw y=x+2; intersections give \(\boxed{x\approx-3.86,-1.32,1.18}\).

(a) \(\boxed{dy/dx=2-2x}\). (b)(i) At x=−1, gradient=2−2(−1)=4. Through A, y−0=4(x+1), so y=4x+4. (ii) Perpendicular gradient=−1/4. Through B, \(y-3=-\frac14(x-2)\), hence \(\boxed{y=-x/4+7/2}\). (c) Set 2−2x=0, givingx=1. Then y=4. Since second derivative−2<0, the maximum is \(\boxed{(1,4)}\).

(a) \(f^{\prime}(x)=3x^2-6x\). At \(x=1\), gradient \(=3-6=-3\).

(b) \(3x(x-2)=0\) gives \(x=0,2\). Turning points \((0,-4),(2,-8)\). Since \(f^{\prime\prime}(x)=6x-6\), they are a maximum and minimum respectively.

(c) A positive cubic rising from bottom left to top right, with both turning points below the \(x\)-axis and the local maximum on the \(y\)-axis.

(a) At \(x=-0.5\), \(y=0.75\); at \(x=0.5\), \(y=-1.25\).

(b) Plot all six points and join with a smooth parabola.

(c) Rewrite \(3x^2-4x-2=0\) as \(3x^2-2x-1=2x+1\). Draw \(y=2x+1\). Its intersection gives \(x\approx-0.39\) (about \(-0.4\) from the graph). The other root \(1.72076\ldots\) is outside the interval.

(a) \(\frac{dy}{dx}=3x^2-12\). At \(x=1\), gradient \(=-9\).

(b) \(3x^2-12=0\Rightarrow x=\pm2\). Substituting gives \((-2,16),(2,-16)\).

(a)(i) \((x-4)(x+1)=x^2-3x-4\). Multiply by \(x-2\): \(x^3-3x^2-4x-2x^2+6x+8=x^3-5x^2+2x+8\).

(ii) A positive cubic crosses the x-axis at \((-1,0),(2,0),(4,0)\) and y-axis at \((0,8)\).

(b) \(\frac{dy}{dx}=3x^2-10x+2\). Set equal to 10: \(3x^2-10x-8=(3x+2)(x-4)=0\). Thus \(x=-2/3,4\). Corresponding points \((-2/3,112/27),(4,0)\).

Using \(y-y_0=10(x-x_0)\), the equations are \(y=10x+\frac{292}{27}\) and \(y=10x-40\).

(a) At \(x=-1,0,1\) the values are \(-2.5,-2,-1\). (b) Plot all eight points and join with a smooth increasing exponential curve.

(c) Intersection with \(y=2\) gives \(x\approx2.32\) (exactly \(log_2 5\)).

(d) Rearrange \(2^x-x-1.5=0\Rightarrow2^x-3=x-1.5\). Draw \(y=x-1.5\). The intersection x-values are \(-1\text{ and approximately }1.66\).

(a) \(\frac{dy}{dx}=4x^3-16x\). (b) Set derivative to zero: \(4x(x^2-4)=4x(x-2)(x+2)=0\). Thus \(x=0,-2,2\). Substitution gives \((0,5),(-2,-11),(2,-11)\).

(c) \(\frac{d^2y}{dx^2}=12x^2-16\). At \(x=0\) this is \(-16<0\), so \((0,5)\) is the maximum. At \(x=\pm2\) the second derivative is 32, so those points are minima.

(a) Cubic: the graph has two turning points and opposite end behaviours.

(b)(i) Two reciprocal branches in quadrants I and III. Both axes are asymptotes; neither branch crosses an axis.

(ii) Multiply by \(2x\): \(1=4x^2\Rightarrow x=\pm1/2\).

(c)(i) Sine passes through \((0,0),(90,1),(180,0),(270,-1),(360,0)\).

(ii) \(\sin x=-1/3\). Reference angle \(\alpha=\sin^{-1}(1/3)=19.4712\ldots^\circ\). Solutions \(x=180+\alpha=199.5^\circ\) and \(x=360-\alpha=340.5^\circ\).

Differentiating gives \(14ax^6+3kx^{k-1}\). Compare coefficients: \(14a=42,3k=15\), so \(a=3,k=5\).

(a) \(\frac{dy}{dx}=3x^2-8x-3\). (b) \(3x^2-8x-3=(3x+1)(x-3)=0\). Thus \(x=-1/3,3\). Substitute into the curve to obtain \((-1/3,14/27),(3,-18)\).

(c) \(\frac{d^2y}{dx^2}=6x-8\). At \(x=-1/3\) this is \(-10<0\), so \((-1/3,14/27)\) is a maximum. At \(x=3\) it is \(10>0\), so \((3,-18)\) is a minimum.

(a) \(dy/dx=12x^2-4x^3\). (b) \(4x^2(3-x)=0\) gives \(x=0,3\). The maximum is at \(x=3\), \(y=4(27)-81=27\), so \(B=(3,27)\).

(c) \(x^3(4-x)=0\) gives the other intercept \(x=4\). Gradient \(12(4)^2-4(4)^3=-64\).

(a) \(18+5x-2x^2=(x+2)(9-2x)\). Thus \(A=(-2,0),B=(0,18),C=(4.5,0)\). (b) \(dy/dx=5-4x\).

(c) \(5-4x=17\) gives \(x=-3\). Then \(y=18+5(-3)-2(-3)^2=-15\). The point is \((-3,-15)\).

(a) \(\tan^{-1}(11.43)=85.0^\circ\). Tangent is positive in quadrants I and III, so \(x=85,265\) degrees. (b) \(y=x(2-x)(2+x)\) crosses at \(x=-2,0,2\); it rises to the left and falls to the right.

(c) \(dy/dx=3x^2+a\). At \(x=2\), \(12+a=0\) so \(a=-12\). Then \(k=8-24+b=b-16\) and \(10-k=-8+24+b=b+16\). Adding gives \(10=2b\), so \(b=5,k=-11\).

(a) At \(x=-0.75,-0.5\) the values are \(1.0,0.9\). (b) Draw a smooth negative branch through the table points.

(c) Draw \(y=-1\). The intersection is approximately \(x=0.31\). The negative branch has no intersection with this line.

(a)(i) −3. (ii) \(x=-1,1.6,4.4\) approximately. (iii) Minimum is just below −8; the smallest integer strictly above it is \(k=-8\). (iv) One possible line is \(y=-5x\); it meets the given curve only once in the stated interval.

(b)(i) \(dy/dx=6x-12\), at \(x=5\) giving 18. (ii) \(6x-12=0\) gives \(x=2\), \(y=-5\), so \((2,-5)\). (c) \(dy/dx=2px^{p-1}+2qx\). Match powers and coefficients: \(p=7,q=3\).

(a) \(\frac{dy}{dx}=3x^2-2kx\). Substitute the gradient and x-value: \(6=12-4k\), so \(k=1.5\). (b) \(3x^2-3x=3x(x-1)=0\). At \(x=0\), \(y=1\); at \(x=1\), \(y=0.5\). Points \((0,1),(1,0.5)\).

(c) Positive cubic, maximum on the positive y-axis at \((0,1)\) and minimum at \((1,0.5)\). Both stationary points are above the x-axis; it crosses the x-axis once to the left of zero.

(a) Negative cubic with x-intercepts \((-3,0),(-1,0),(3,0)\) and y-intercept \((0,9)\).

(b)(i) \((3-x)(3+x)=9-x^2\). Thus \((x+1)(9-x^2)=9+9x-x^2-x^3\). (ii) \(\frac{dy}{dx}=9-2x-3x^2=0\). \(x=\frac{-2\pm\sqrt{4+108}}6=\frac{-1\pm\sqrt{28}}3\). Turning x-values \(-2.10,1.43\).

(iii) Using unrounded x-values, local minimum \(-5.049\ldots\) and local maximum \(16.900\ldots\). Below the minimum or above the maximum there is one intersection. Required integers: \(a=-6,b=17\).

Maximum at the origin, minimum in the fourth quadrant with x-coordinate 4; rising before 0, falling from 0 to 4, rising after 4. One suitable example is \(y=x^3-6x^2\).

(a) At x = 0, \((-3)b(2)=-30\) gives \(b=5\). (b) Roots are 3, −5, −2. The positive root gives \((3,0)\).

(a)(i) One choice is k = 4, giving one crossing on each side of the minimum at D. (ii) Graph range is from 3 to 10 inclusive. Thus \(a=3,b=10\). (b) \(\frac{dy}{dx}=6x^5-30x^4=6x^4(x-5)\). Set this to zero: x = 0 or 5. Corresponding y-values are 0 and \(5^6-6(5^5)=-3125\). Points \((0,0),(5,-3125)\).

(a)(i) \(2^2(9-2)=28\). (ii) Plot the table points and join with a smooth curve.

(iii) Intersections with V = 44 give x approximately 2.6 or 8.4. (b)(i) Two square ends and four side faces give \(S=2x^2+4x(9-x)=2x^2+36x-4x^2=36x-2x^2\). (ii) Maximum volume is at x = 6, from the graph (or \(V\prime=18x-3x^2=0\)). Then \(S=36(6)-2(6^2)=144\text{ cm}^2\).

(a)(i) \(y=3x/4-3\) gives (0, −3) and (4, 0). (ii) \(y=(x-4)(x+1)\) gives (−1, 0), (4, 0), and (0, −4); minimum at (1.5, −6.25). (iii) Increasing exponential, intercept (0, 1), no x-intercept, horizontal asymptote y = 0.

(b)(i) \(\frac{dy}{dx}=8-4x^2\). (ii) \(8-4(-1)^2=4\). (iii) \(8-4x^2=-28\) gives \(x^2=9\) and x = ±3. At x = 3, y = −7; at x = −3, y = 17. Points (3, −7) and (−3, 17).

(a) Rectangular hyperbola in quadrants I and III, asymptotes x = 0 and y = 0. (b) Increasing exponential through (0, 1); horizontal asymptote y = 0, with y always positive.

\(\frac{dy}{dx}=3x^2-4x=x(3x-4)\). Set to zero: \(x=0,4/3\). At x = 0, y = 5. At x = 4/3, \(y=64/27-32/9+5=103/27\). Points \((0,5),(\frac43,\frac{103}{27})\).

(a)(i) At x = 0, a = 4. At (1, 1), \(1=4+b\) gives b = −3. (ii) Horizontal tangent y = 4. (iii) Symmetry in the y-axis changes gradient 6 to −6; through (1, 1), \(y-1=-6(x-1)\), so y = −6x + 7.

(b)(i) Missing values: 2.25, 2.67, 3.5. (ii) Plot the completed table and join smoothly.

(c)(i) Approximately x = −0.74 and 0.56. (ii) \(4-3x^2=1+\frac5{3-x}\). Multiply by 3 − x: \((3-3x^2)(3-x)=5\). Expand and rearrange: \(3x^3-9x^2-3x+4=0\).

\(\frac{dy}{dx}=5x^4-20x^3=5x^3(x-4)\). Set this to zero: \(x=0,4\).

(a) Hyperbola in quadrants I and III, with both axes as asymptotes. (b) Decreasing exponential through (0, 1), always positive, approaching the x-axis as x increases.

(a) Completed y-values: −2, 0.75, 3, 6, 7, 6, 3, 0.75, −2. (b) Downward parabola with vertex (2, 7). (c) For example k = 8, above its maximum of 7. (d) Rewrite as \(3+4x-x^2=4-\tfrac12x\). Draw \(y=4-x/2\) and read intersections: approximately x = 0.23 and x = 4.27. Algebra confirms \(x=(9\pm\sqrt{65})/4\).

(a) \(\frac{(x-5)(x+5)}{(x-5)(x+4)}=\frac{x+5}{x+4}\), with \(x\ne5,-4\). (b) \(\frac{(x+5)(x-1)+x(x+8)}{x(x-1)}=\frac{2x^2+12x-5}{x(x-1)}\), \(x\ne0,1\). (c)(i) \(dy/dx=6x^2-8x\). (ii) \(6(4^2)-8(4)=64\). (iii) \(2x(3x-4)=0\) gives \(x=0,4/3\). Substituting gives points \((0,6),(4/3,98/27)\).

(a)(i) At x = 0, y = 1; at x = 2, y = −0.5. (ii) Draw a decreasing exponential curve through the table points. (b) Rewrite as \(y=2.5-2x\); the intersection is approximately x = 1.36. (c) Since \(2(.5)^x>0\) for every real x, y is always greater than −1. Highest excluded integer k = −1; asymptote \(y=-1\).

(a)(i) \(x^3-4x^2+4x=x(x-2)^2\), so a = 2. (ii) A positive cubic crosses the axes at (0, 0) and touches the x-axis at (2, 0). (b) \(dy/dx=3x^2-8x+4\). At x = 4, gradient \(48-32+4=20\) and y = 16. Tangent \(y-16=20(x-4)\), hence \(y=20x-64\).

(a)(i) The missing values are 9.5, 4.8 and 8.5. (ii) Draw two smooth separate branches, leaving x = 0 undefined. (b) Draw \(y=4.8-2x\). Its two intersections within the stated intervals give approximately \(x=-.28\) and \(x=1.60\). (c) Multiply by 10x: \(10x^3-15=48x-20x^2\), so \(10x^3+20x^2-48x-15=0\). Hence \(a=10,b=20,c=-48\).

(a) \(\frac{dy}{dx}=3x^2-12=0\) gives \(x=\pm2\). Substitution gives (−2, 22) and (2, −10). (b) \(\frac{d^2y}{dx^2}=6x\). At x = −2 it is −12, so (−2, 22) is a maximum. At x = 2 it is 12, so (2, −10) is a minimum.

(a) Missing y-values: −1, −0.375, 3. (b) Plot all nine points and draw a smooth cubic curve. (c) Rearrange to \(x^3-3x^2+3=2-x\), so draw \(y=2-x\). Intersections give \(x\approx-.41,1,2.41\).

(a) (0, 0), (1, 0), (2, 0). (b) \(x(x^2-3x+2)=x^3-3x^2+2x\). (c) \(f\prime(x)=3x^2-6x+2=0\). \(x=\frac{6\pm\sqrt{36-24}}6=1\pm\frac1{\sqrt3}\). Substitution gives \((1-1/\sqrt3,\ 2/(3\sqrt3))\) and \((1+1/\sqrt3,\ -2/(3\sqrt3))\). To 1 decimal place: (0.4, 0.4) and (1.6, −0.4). (d) Cubic rises from bottom left, crosses 0, reaches its local maximum, crosses 1, reaches its local minimum, crosses 2 and rises to the right.

(a) Straight line of gradient 1 through (0, −3) and (3, 0). (b) Two hyperbola branches in quadrants I and III, with the coordinate axes as asymptotes.

\(\frac{dy}{dx}=3x^2-3=0\) gives x = ±1. Points are (−1, 6) and (1, 2). The second derivative is \(6x\): at x = −1 it is −6, so (−1, 6) is a maximum; at x = 1 it is 6, so (1, 2) is a minimum.

(a)(i) Missing values: 3, 2.25, 1. (ii) Plot the table points and join with a smooth cubic. (iii) Intersections with y = 1.5 give approximately \(x=-.54,.79,1.76\). (iv) One choice is k = 0: the horizontal line meets the displayed curve once.

(b) A tangent through about (1, 3) and (0, −2) gives \(m\approx(3-(-2))/(1-0)=5\). Its equation is approximately \(y=5x-2\).

At an intersection, \(x^2-2x+1=mx+c\). Rearranging gives \(x^2-(2+m)x+(1-c)=0\). Compare coefficients: m = 5, c = −4. Thus \(y=5x-4\).

(a)(i) \(AC=\sqrt{(4+3)^2+(-4-10)^2}=\sqrt{245}=15.7\). (ii) Gradient \((-4-10)/(4+3)=-2\). Substitute A: \(10=-2(-3)+c\), so c = 4.

(iii) Rhombus diagonals bisect each other at right angles. Midpoint of AC is (1/2,3), and BD has gradient1/2. Thus \(y-3=\frac12(x-\frac12)\Rightarrow y=\frac12x+\frac{11}4\).

(b)(i) \(\frac{dy}{dx}=3x^2+16x+5=(3x+1)(x+5)\). Set to zero: x = −1/3 or −5. Substitute into y: turning points \((-\frac13,-\frac{22}{27})\) and (−5,50). (ii) \(\frac{d^2y}{dx^2}=6x+16\). At x = −1/3 this is14 > 0: minimum. At x = −5 it is−14 < 0: maximum.

(a)(i) \((-1)^4-4(-1)^3=5\). (ii) \(\frac{dy}{dx}=4x^3-12x^2=4x^2(x-3)\). Set to zero: x=0 or3. Corresponding y values0 and−27. Stationary points(0,0) and(3,−27).

(b) Differentiate: \(px^{p-1}+2qx^{q-1}=11x^{10}+10x^4\). Matching powers and coefficients gives p=11 and q=5.

(a)(i) \(f(x)=x+5\). The intercepts are(−5,0) and(0,5), giving gradient1. (ii) \(f(x)=2\sin x\) with x in degrees: amplitude2 and period360°.

(b) Draw a tangent at P. Using two well-separated points on it, approximately(1,1.36) and(2.4,3.26), gives \(\text{gradient}\approx\frac{3.26-1.36}{2.4-1}\approx1.36\).

(a)(i) \(\frac{dy}{dx}=12x^2-3=0\) gives x=±1/2. Substitution gives(−1/2,4) and(1/2,2). (ii) \(\frac{d^2y}{dx^2}=24x\). At−1/2 this is−12, a maximum; at1/2 it is12, a minimum.

(b) Rewrite as \(x^2-x+1=x+3\). Draw the line y=x+3; its intersections give x≈−0.73 and2.73 (within the graph-reading accuracy).

(a) \(\frac{dy}{dx}=4-2x\). (b) Set4−2x=0: x=2. Substitute: y=6+8−4=10. Turning point(2,10).

(a)(i) Factor \(x^2+3x-4=(x+4)(x-1)\). Thus A=(−4,0), B=(1,0); at x=0, C=(0,−4). (ii) \(dy/dx=2x+3\). (iii) Gradient at x=2 is7, so \(y-6=7(x-2)\Rightarrow y=7x-8\).

(b)(i) Tangent has zeros0°,180°,360°, vertical asymptotes90° and270°, and increasing branches. (ii) \(\tan x=-7/5\). Reference angle \(\alpha=\tan^{-1}(7/5)=54.462\ldots^\circ\). Therefore \(x=180^\circ-\alpha=125.5^\circ\) or \(360^\circ-\alpha=305.5^\circ\).

(a)(i) Read where the horizontal line y=14 meets the curve: x≈2.75. (ii) A tangent at (−2,4) gives, for example, rise 8 over run 1, so gradient≈8. (iii) Draw y=2x−2; its intersection with the curve in the stated interval gives x≈−2.85.

(b) \(2x^2-2x-7=3x+5\Rightarrow2x^2-5x-12=(2x+3)(x-4)=0\). Thus x=4 or −1.5. Substitute into y=3x+5: A=(4,17), B=(−1.5,0.5).

(a) At x=1, y=2; at x=2, y=4.5. (b) Plot the table values and join with a smooth curve; its minimum is near (0.794,1.890). (c)(i) Draw y=−2: x≈−0.45. (ii) Draw y=1−x: x≈−1.84. (d) For example k=3: the line y=3 crosses the negative branch once and the positive branch twice. Any integer greater than2 works.

(a)(i) \((8-x)(x+3)\). (ii) Roots−3 and8 give a=−3,b=8; y-intercept c=24. (iii) \(dy/dx=5-2x\), so at−1.5 the gradient is8.

(b)(i) Positive cubic: crosses at(−1,0), passes through(0,9), touches the x-axis at the local minimum(3,0), then rises. (ii) \((x+1)(x^2-6x+9)=x^3-5x^2+3x+9\).

Draw a ruled tangent touching the curve at P≈(2,1). A suitable tangent passes through approximately(1,1.5) and(3,0.5), giving \((0.5-1.5)/(3-1)=-.5\).

(a) Missing values, in order:−2.1,1.6,−1.7,2.1. (b) Plot all nine points and join smoothly. (c) Draw \(y=(1-x)/2\). The intersections give x≈−2.05,−0.34,2.39. (d) The horizontal line y=1 crosses the curve three times: n=3.

(a) At x=−3,0,1 the values are 2,2,6. (b) Plot the points and draw a smooth cubic. (c) The x-axis crossing is x≈−3.20. (d) Draw y=−2x; its crossing with the cubic gives x≈−2.52. (e) Turning levels are 2 and 6, so k=3,4 or 5 gives three crossings.

(a) Missing values at x=.2,1,2 are2.45,.25,−.25. (b) Join plotted points with a smooth curve. (c) Read where y=.5: x≈.73. (d) Intersections with y=2−x are x≈.28 and2.39. (e) At y=0, \(\frac1{2x}-\frac x4=0\Rightarrow x^2=2\). The positive x-axis crossing is \(x=\sqrt2\approx1.41\).

(a)(i) f(1)=−3. (ii) f(−2)=3, then f(3)=19/3≈6.33. (b) Since \(f(x)=x^2-\frac2x-2\), draw y=5−3x. Crossings within −3≤x≤3 give x≈−.26 and1.73; another algebraic root≈−4.47 lies outside the interval.

(c) A tangent at(−2,3) passes through approximately(−3,6.5),(−1,−.5), so gradient≈−3.5. (d)(i) Missing values8,4,.25. (ii) Plot and draw a smooth decreasing exponential. (iii) The positive crossing is x≈1.84.

(a) At x=−1, y=3.5; at x=.2, y=15; at x=3, y=3.9 (1 decimal place). (b) Plot both branches, keeping them separate across x=0.

(c) The positive branch is at or below the x-axis between about x=.52 and x=1.36: \(.52\le x\le1.36\). (d) k=1 gives two intersections on the positive branch; it is the smallest positive integer. (e)(i) Draw y=3x+1; the intersection in the plotted intervals is x≈.36. (ii) Multiply by 2x²: \(x^4+2-4x=6x^3+2x^2\Rightarrow x^4-6x^3-2x^2-4x+2=0\). Thus a=−6, b=−2, c=−4.

(a) Missing values are 5, −3, 21. (b) Plot the points and draw a smooth curve.

(c) Intercepts approximately x=−2.79, 0, 1.79. (d) The tangent at (2,2) passes through (1,−9) and (3,13), so gradient \((13-(-9))/(3-1)=11\). (e) The local maximum is between 6 and 7, so the largest integer is k=6.

(a) At .5, y=0; at .75, y=−.17; at 2.5, y=2.4. (b) Draw a smooth curve through the plotted values.

(c) Read intersections with y=2, at about .22 and 2.28. Within the given interval, \(.125\le x\lesssim.22\ \text{or}\ 2.28\lesssim x\le3\). (d)(i) Substitute \(1/x=7-3x\) into y=2x+1/x−3 to get y=4−x. (ii) Intersections give x≈.15 and x≈2.18.

(a) x=0. (b) At (0.5,8), gradient −8; tangent y=−8x+12. (c)(i) Missing entries: 0, 2.375≈2.4, 4. (ii) The smooth cubic is drawn below. (d) \(x^3+3x+4=10-8x^2\Rightarrow x^3+8x^2+3x-6=0\). (e) Rearrange to \(x^3+3x+4=-2x+2\). Draw y=−2x+2; its intersection with the cubic gives x≈−0.39 (about −0.4 from the graph).

(a) Missing values: −2.0417≈−2.0, −0.1667≈−0.2, 2.5417≈2.5. (b) Draw the two smooth branches, without joining across x=0. (c)(i) A tangent gives gradient about 3.9 (exactly 31/8). (ii) At x=−2, y=−67/24, so \(y=\frac{31}8x+\frac{119}{24}\approx3.875x+4.958\). (d)(i) Intersection with y=0 gives x≈1.08. (ii) Intersections with y=−4 give x≈−2.27, −0.35, 0.35. (e) Multiply by 6x²: \(2x^5+24x^2-3=0\). Thus a=2, b=24, n=5.

(a)(i) 1 and16. (ii) 14 and−2. (b) Both smooth curves are drawn below. (c)(i) x≈3.58 (about3.6 from the graph). (ii) x≈2.72. (d)(i) \(y-2=-4(x-4)\Rightarrow y=-4x+18\). (ii) Tangent at (2,10): substituting the line gives \(14-x^2=-4x+18\Rightarrow(x-2)^2=0\), a double intersection.

(a) Missing y values 0,−2,0.9. (b) See curve. (c) Approximately−0.41<x<1 and x>2.41 within the drawn domain. (d)(i) Rearrange as y=1−x. (ii) Intersection gives x≈2.32. (e) Tangent gradient≈2.69; analytical check: \(\frac{dy}{dx}=3x^2-6x+1\) gives 2.6875 at−.25.

(a) Missing values−3 and 17. (b) See curve. (c)(i),(ii) The tangent touches at(2,1). (iii) Gradient=(1+17)/2=9; y=9 x−17. Its gradient also equals 3(2)²−3=9. (d) Draw y=3 x+2, since \(x^3-3x-1=3x+2\). Intersection estimates x≈−2.15,−.52,2.67.

(a)−1 and 3. (b) See graph. (c) Tangent slope≈−1.75; analytical check \(f^{\prime}(x)=x/2+4/x^2\) gives−1.75. (d)−3 and 3. (e) See curves. (f)(i)x≈3.73. (ii)x>3.73. (g) \(\frac{x^2}4-\frac4x=\frac9x\Rightarrow x^3=52\), so k=52.

(a) At x=0,1,2, the values are 0,0.5,1.25. (b) Join the plotted values with a smooth increasing exponential curve. (c) \(x=\log(4.5)/\log(1.5)=3.71\ldots\), approximately 3.7 from the graph. (d) Draw \(y=x+1\). Intersections give x≈−1.44 and 4.69. (e) A is(5,5). Draw the tangent through A and calculate its gradient using two points on that tangent; it is about 1.30.

(a) Read y at x=1: approximately −1.5. (b) Read x at y=3: −0.5. (c) \(k>-4\) gives one intersection. Because the domain is finite, \(k<f(-2.5)\) (approximately \(k<-9.6\)) also gives one intersection, on the left branch near the vertical asymptote. (d) Draw y=x−5; intersections give x≈−2.2,−1.15,1.35. (e) A tangent at (1,−1.5) gives a gradient of about −5.2, using change in y divided by change in x.

(a) \(2x^2-1=5\Rightarrow x=\pm\sqrt3\approx\pm1.73\). (b) Tangent is \(y-3.5=-6(x+1.5)\); gradient≈−6. (c)(i) g(0)=1. (ii) Plot the table points and join smoothly.

(d)(i) Intersections are approximately −0.8786 and 1.3233, so graph readings −0.88 and 1.32. (ii) \(-0.88<x<1.32\) approximately. (e)(i) \(1/8=0.125,\ 1/32=0.03125,\ 1/1024=0.0009765625\). (ii) 0.

(a) Missing y values: 0,2.25,2,1.25, by substitution. (b) Plot all points and join with a smooth curve.

(c) 1: the local maximum at x=−4/3 has y=64/27≈2.37<3, so only the rising right branch reaches 3. (d)(i) y=x+1. (ii) x≈−2.12,−0.43,0.55. (e) \(\frac{dy}{dx}=6x^2+8x=2x(3x+4)\) is negative for \(-4/3<k<0\).

(a) f(−1)=10,f(2)=7. (b) Plot the table and join with a smooth curve.

(c) x≈−1.63. (d) Tangent at (3.5,8.875) has gradient≈8.75; \(y-8.875=8.75(x-3.5)\). (e) Rewrite \(f(x)=2x+10\) and draw y=2x+10. Intersections give x≈−1.193,1,4.193.

(a) y(3)≈3.2,y(3.5)≈5.2. (b) Plot the positive branch points, joining smoothly.

(c) x≈1.74. (d)(i) For example k=0. (ii) k=−2 gives three intersections. (e) \(\frac{dy}{dx}=3x^2/8+4/x^3\), which is about 3.23 at x=−3; draw the corresponding tangent. (f)(i) Draw y=6−x: intersection x≈2.97. (ii) Multiplying by 8x² gives \(x^5-16=48x^2-8x^3\), so a=8,b=−48,c=−16.

(a) The missing values are \(3.50,1.94,3.11\). (b) Plot the points and join smoothly in two separate branches. (c) Read the intersection with \(y=1.5\): \(x\approx-0.68\). (d)(i) Add \(x\) to both sides: \(x+\frac1{x^2}=2.5-x\). Thus \(a=-1\) and \(b=2.5\). (ii) Read the intersection with \(y=2.5-x\): \(x\approx-0.53\). (e) A tangent at \((2,2.25)\) gives a gradient of approximately 0.75. For example, the tangent passes through \((1,1.5)\) and \((3,3)\), so its gradient is \((3-1.5)/(3-1)=0.75\).

(a) \(f(5)=20/5+5=9\) and \(f(8)=20/8+8=10.5\). (b) Draw the two smooth branches through the points. (c) The intersections with \(y=11\) give \(x\approx2.3\) or \(x\approx8.7\). For verification, \(x^2-11x+20=0\Rightarrow x=\frac{11\pm\sqrt{41}}2\). (d) For positive \(x\), \(x+20/x\ge2\sqrt{20}\approx8.944\); for negative \(x\) the values are negative. Thus the positive primes with no solution are 2, 3, 5 and 7. (e) The opposite point is \((-2,-12)\); \(f(-x)=-f(x)\) gives equal gradients at opposite points.

(f)(i) \(20/x+x=x^2\Rightarrow20+x^2=x^3\Rightarrow x^3-x^2-20=0\). Thus \(p=-1,q=-20\). (ii) Draw the parabola through \((0,0),(\pm1,1),(\pm2,4),(\pm3,9),(\pm4,16)\). (iii) The intersection gives \(x\approx3.1\) (the root is \(3.091978\ldots\)). (iv) The same equation defines the x-intercept of the cubic, so \(n\approx3.1\).

(a)(i) \(m=\frac{2-0}{0-4}=-\frac12\) and the y-intercept is 2, so \(y=-\frac12x+2\). (ii) Substituting A gives \(16/a^2=1\Rightarrow a^2=16\). Substituting B gives \(4/b^2=1\Rightarrow b^2=4\).

(b)(i) \(\frac{2^2}{16}+\frac{k^2}{4}=1\Rightarrow\frac{k^2}{4}=\frac34\Rightarrow k=\sqrt3\). (ii) The angle POQ is twice the angle between OP and the positive x-axis: \(2\tan^{-1}(\sqrt3/2)=81.7868\ldots^\circ\approx81.8^\circ\). (c)(i) \(\pi(4)(2)=8\pi\). (ii) The length scale factor is \(12/4=3\). Area scales by \(3^2=9\), giving \(72\pi\).

(a)(i) \(f(-1)=1+1-4=-2\) and \(f(2)=4-1/2-4=-0.5\). (ii) Plot the two branches smoothly through the table points. (b) The x-intercepts are approximately \(x=-1.86,-0.25,2.11\). (c) One example is \(k=-3\). The negative branch stays above −3, while the positive branch increases from below −3 to above it. (d)(i) Draw \(y=-5x-2\). The two intersections in the specified intervals give \(x\approx-0.30\) and \(x\approx0.63\).

(d)(ii) \(x^2-\frac1x-4+2=-5x\Rightarrow x^2+5x-2-\frac1x=0\). Multiplying by \(x\) gives \(x^3+5x^2-2x-1=0\). Thus \(a=5,b=-2\).

(a)(i) Read \(f(-2)=10\). (ii) The intersections with the horizontal line \(y=2\) give approximately \(x=-3.35,-0.35,1.65\). (iii) The horizontal tangents are at the maximum and minimum, giving \(y\approx10.0\) and \(y\approx-2.2\). (b)(i) \(g(-2)=2\), \(g(-0.5)=-1\) and \(g(2)=4\). (ii) Draw the two branches through the table points. (iii) The intersections give approximately \(x=-3.3\) and \(x=1.85\).

(c) \(g(f(-2))=g(10)=2/10+3=3.2\). (d) Solve \(g(x)=5\): \(2/x+3=5\Rightarrow x=1\). Thus \(g^{-1}(5)=1\).

(a) At \(x=1\), \(y=-1\). Draw a tangent there. A suitable tangent passes through \((0,-4)\) and \((2,2)\), giving gradient \((2-(-4))/(2-0)=3\). (b) Read the maximum point \(D=(-2,8)\).

(a) The missing values, in order, are 0.92, 0.5, −1, −1, 0.5 and 0.92. (b) Draw two smooth branches through the table points. (c)(i) The line passes through (−2, 1) and (2, −3). (ii) Read \(x\approx0.84\). (iii) Multiply by \(x^2\) and rearrange: \(x^2-2=-x^3-x^2\Rightarrow x^3+2x^2-2=0\). Hence p = 2, q = −2. (d)(i) \(1-2/x^2=0\Rightarrow A=(\sqrt2,0)\approx(1.41,0)\). (ii) The line has equation \(y=\sqrt2x-2\). (iii) It is a tangent to the curve at A.

(a) Both missing values are 1. (b) Draw a smooth cubic through all points. (c)(i) Intersections with the x-axis give approximately \(x=-0.88,1.35,2.53\). (ii) Draw \(y=-x\) and read the intersection with the curve: \(x\approx-0.59\). (d)(i) The horizontal tangents are \(y=1\) and \(y=-1/3\). (ii) The minimum y is \(-1/3\approx-0.33\).

(a) At −2, \(-6+2/4+1=-4.5\). At 0.5, \(1.5+2/0.25+1=10.5\). (b) Draw the two smooth branches shown. (c) The positive branch has minimum approximately 5.95, while the negative branch meets each horizontal line once. The largest integer below 5.95 is \(k=5\). (d)(i) Draw \(y=15-3x\). Intersections give approximately \(x=-0.35,\ 0.42,\ 2.27\). (ii) Multiply by \(x^2\) and rearrange: \(6x^3-14x^2+2=0\). Thus \(a=6,\ b=-14,\ c=0\).

(a) Missing values, in order: 2, 0, −2, 2. (b) Draw a smooth cubic through the values. (c) Draw \(y=\frac12(x+1)\). Intersections give \(x\approx-2.87,\ -1,\ 0.87\). (d) A tangent at x = −1.75 gives gradient approximately −1.3. This is consistent with \(3x^2+6x=-1.3125\) at that point.

(a) Missing values are 1.5, 1.25, −0.75, 0.5. (b) Draw the two smooth branches. (c) Draw y = 2. The intersections give approximately \(x=-1.27,\ -0.26,\ 1.53\). (d) The negative-x branch has minimum approximately 1.2; the positive branch meets every horizontal line once. Thus, from the graph, \(k<1.2\) approximately. The exact threshold is \(k<3/(2^{4/3})=1.19055\ldots\). (e) The tangent gradient is approximately −1.5, consistent with \(2x+1/(2x^2)=-1.5\) at x = −1.

(a) Missing values: 2, 7. (b) Draw two smooth branches, leaving x = 0 undefined. (c) The tangent gradient is approximately −12; checking \(2x-2-12/x^2\) at x = 1 gives −12. (d) The required horizontal line touches the positive branch at its minimum and crosses the negative branch. Approximately \(k=5.89\), with \(x=2.22\) and \(x=-2.44\). (e) At the line intersections, \(x^2-2x+12/x=3x+1\). Multiply by x and collect terms: \(x^3-5x^2-x+12=0\), giving \(a=-5,\ b=-1,\ c=12\).

(a) Missing values: 1, 3, 2.5. (b) Draw both smooth branches. (c) The x-intercept is approximately \(x=-2.52\); checking algebra gives \(x^3=-16\). (d) Draw y = 1 − x. Intersection is approximately \(x=-1.55\), consistent with \(3x^3-2x^2+16=0\). (e) Tangent gradient is approximately 1.1. Checking \(f^{\prime}(x)=-16/x^3+1/2\) at −3 gives \(16/27+1/2=59/54\approx1.09\).

(a) \(2^{0.7}=1.6245\ldots\approx1.62\). (b)(i) 7. (ii) \(x+9=13\Rightarrow x=4\). (iii) \(9-x=2\Rightarrow x=7\). (iv) \(x=1/3\). (c)(i) Missing values 0.25 and 1. (ii) Draw a smooth increasing curve. (iii) Approximately \(x=2.3\) (exactly \(\log_2 5=2.3219\ldots\)). (iv) Gradient \((8-2)/(3-1)=3\); using (1, 2) gives \(y=3x-1\). (v) Draw y = x + 2. Intersections approximately \(x=-1.7\) and \(x=2\).

(a) Missing y values: 0, 4, 0.625, 0.875. (b) Plot all table points and join smoothly. (c) Draw y = x + 1. Read intersections approximately 0.25 and 1.86. The third root lies outside the given interval. (d) Draw the tangent at (−1.5, 3.125); its gradient is approximately 3.75. This agrees with \(dy/dx=3x^2-3\) at x = −1.5.

(a) −1.5 and 0.5. (b) Plot both branches smoothly without joining across x = 0. (c) Approximately x = 1.30. (d) k = −1. The negative branch has maximum −1.5 at x = −1, while the positive branch is increasing through all real y values. Therefore a horizontal line y = −1 meets only the positive branch; every other negative integer meets both branches. (e)(i) Rearrange to \(x-1/(2x^2)=2-x\), so draw y = 2 − x. (ii) Its intersection is approximately x = 1.18.

(a) 2.125, 2.375. (b) Plot and join smoothly. (c) Tangent gradient approximately 9, consistent with \(3x^2-3\) at x = 2. (d) \(x(x^2-3)=0\), so approximately −1.73, 0, 1.73. (e) Turning-point y values are 3 and −1, hence −1 < k < 3.

(a) 3, 3, −1. (b) Plot both branches smoothly, retaining the gap at x = 0. (c) Approximately 0.57, by intersecting y = 2. (d) Draw y = 2x + 3; intersection approximately 0.46. (e) Draw a tangent at (−1.5, 3.44…). Its gradient is approximately −1.41, consistent with \(f^\prime(x)=-2/x^3-2\).

(a) 0, 4.5, 3.11. (b) Draw smooth branches without joining across x = 0. (c) Intersections with y = 3 give approximately −0.53, 0.65, 2.88. (d) Draw y = 1 − x; intersection approximately −0.66. (e) Tangent gradient approximately 0.75, consistent with \(y^\prime=1-2/x^3\). (f) \(0=1/x^2+x\Rightarrow0=1+x^3\Rightarrow x^3=-1\Rightarrow x=\sqrt[3]{-1}=-1\).

(a) −2, 5.5. (b) Draw separate smooth branches. (c) Intersections with y = 5: approximately −2.49, 0.66, 1.83. (d) Draw y = x + 5; intersections approximately −2.09, 0.57, 2.51.

(a) −3, 7.375, 8.875. (b) Draw a smooth cubic curve. (c)(i) For example k = 0, below the local minimum near 6.97. (ii) For example k = 8, between the local minimum and maximum 10. (d) Draw y = 15x + 2; intersections in the interval approximately −1.39 and 0.46. (e) Gradient approximately 9.75, consistent with \(f^\prime(x)=15x^2-16x\).

(a) Missing values: \(f(0)=3,\ g(-1)=1/3\approx0.33,\ g(0)=1\). (b) Curves are shown below. (c)(i) \(x\approx1.30\). (ii) \(x\approx1.26\). (iii) \(x\approx0.62\). (d) Tangent gradient is approximately −2. The tangent passes through (0.5, 2.25); its equation is \(y=3.25-2x\).

(a) Missing values 0, 2, 0, −3. (b) Curve below. (c) Draw \(y=-1\); intersections \(x\approx1.31,\ 4.19\). (d)(i) \(P\approx(2.65,3.10)\). (ii) \(m\approx0.417\). From the drawn tangent, \(m\approx(3.10-2)/2.65\approx0.415\).

(a) −5.04, 1.75, 0. (b) Draw the two separate branches below. (c) Intersections with y = −3 give \(x\approx-1.53,-0.35,1.88\). (d) Intersections with \(y=2x-2\) give approximately \(x\approx-2.6,-0.38,1\). (e) A tangent at (−2, −4.5) has gradient about 3.75: using points (−3, −8.25) and (−1, −0.75), \(\text{gradient}=(-0.75+8.25)/2=3.75\).