Answer: Stationary point \((-1,-4)\). Intercepts: \((-3,0)\), \((1,0)\), \((0,-3)\).

(a) Complete the square:

\(y=x^2+2x-3.\)

Since

\(x^2+2x=(x+1)^2-1,\)

we get

\(y=(x+1)^2-1-3=(x+1)^2-4.\)

This is a positive quadratic, so the stationary point is a minimum. The minimum occurs when

\(x+1=0,\)

so \(x=-1\), and then \(y=-4\).

Therefore the stationary point is

\((-1,-4).\)

(b) For the \(x\)-intercepts, put \(y=0\):

\(x^2+2x-3=0.\)

Factorise:

\((x+3)(x-1)=0.\)

So the \(x\)-intercepts are

\((-3,0)\quad\text{and}\quad(1,0).\)

For the \(y\)-intercept, put \(x=0\):

\(y=-3.\)

So the \(y\)-intercept is

\((0,-3).\)

The sketch should be an upward-opening parabola passing through these intercepts with minimum point \((-1,-4)\).

Answer: \(3+4x-2x^2=5-2(x-1)^2\), and \(f(x)\leqslant5\).

We start with the main method. Complete the square first, because this immediately shows the turning value and the value of the variable at which it occurs.

First rewrite the quadratic by completing the square:

\(3+4x-2x^2=-2x^2+4x+3.\)

Factor out \(-2\) from the quadratic terms:

\(-2x^2+4x+3=-2(x^2-2x)+3.\)

Since \(x^2-2x=(x-1)^2-1\),

\(-2(x^2-2x)+3=-2\left((x-1)^2-1\right)+3.\)

Therefore

\(3+4x-2x^2=5-2(x-1)^2.\)

Because \((x-1)^2\geqslant0\), the greatest possible value of \(5-2(x-1)^2\) is \(5\).

Hence the range is

\(f(x)\leqslant5.\)

This completes the solution and gives the required result.

Answer: (a) \(-3(x+2)^2+31\); (b) maximum \(31\) when \(x=-2\); (c) \(u=\left(-2+\sqrt{\frac{31}{3}}\right)^2\approx1.48\).

We start with the main method. Complete the square first, because this immediately shows the turning value and the value of the variable at which it occurs.

(a) Start by factorising the quadratic terms:

\(19-12x-3x^2=-3(x^2+4x)+19.\)

Complete the square inside the bracket:

\(x^2+4x=(x+2)^2-4.\)

So

\(19-12x-3x^2=-3\big((x+2)^2-4\big)+19.\)

Hence

\(19-12x-3x^2=-3(x+2)^2+12+19.\)

Therefore

\(19-12x-3x^2=-3(x+2)^2+31.\)

(b) Since \((x+2)^2\geq0\), the term \(-3(x+2)^2\leq0\).

So the greatest possible value occurs when

\((x+2)^2=0,\)

which gives \(x=-2\).

The maximum value is then

\(31.\)

(c) Let \(x=\sqrt u\). Then \(u=x^2\), and the equation becomes

\(19-12x-3x^2=0.\)

Using part (a),

\(-3(x+2)^2+31=0.\)

So

\((x+2)^2=\frac{31}{3}.\)

Thus

\(x=-2\pm\sqrt{\frac{31}{3}}.\)

But \(x=\sqrt u\geq0\), so only the positive value is possible:

\(x=-2+\sqrt{\frac{31}{3}}.\)

Therefore

\(u=x^2=\left(-2+\sqrt{\frac{31}{3}}\right)^2.\)

Numerically,

\(u\approx1.48.\)

This completes the solution and gives the required result.

Answer: (a) \(2\left(x+\frac54\right)^2-\frac18\); (b) \(\left(-\frac54,-\frac18\right)\); (c) \(-\frac94\lt x\lt-\frac14\).

For a quadratic expression, first find its roots or complete the square. The sign of the quadratic then follows from its shape and the position of the roots.

(a) Complete the square:

\(2x^2+5x+3=2\left(x^2+\frac52x\right)+3.\)

Half of \(\frac52\) is \(\frac54\), so

\(2x^2+5x+3 =2\left[\left(x+\frac54\right)^2-\frac{25}{16}\right]+3.\)

Hence

(b) The stationary point is therefore

\(\left(-\frac54,-\frac18\right).\)

(c) Solve

\(2\left(x+\frac54\right)^2-\frac18\lt\frac{15}{8}.\)

Then

\(2\left(x+\frac54\right)^2\lt2,\)

so

\(\left(x+\frac54\right)^2\lt1.\)

Thus

\(-1\lt x+\frac54\lt1.\)

Subtracting \(\frac54\),

\(-\frac94\lt x\lt-\frac14.\)

Answer: (a) \(3\left(x+\frac52\right)^2-\frac{155}{4}\); (b) minimum value \(-\frac{155}{4}\) when \(x=-\frac52\); (c) \(y=1.31\) or \(y=-226\).

Answer: (a) \(3\left(x+\frac52\right)^2-\frac{155}{4}\); (b) minimum value \(-\frac{155}{4}\) when \(x=-\frac52\); (c) \(y=1.31\) or \(y=-226\).

(a) Complete the square:

\(3x^2+15x-20=3(x^2+5x)-20.\)

Now

\(x^2+5x=\left(x+\frac52\right)^2-\frac{25}{4}.\)

Therefore

\(3x^2+15x-20=3\left(x+\frac52\right)^2-\frac{75}{4}-20.\)

So

\(3x^2+15x-20=3\left(x+\frac52\right)^2-\frac{155}{4}.\)

(b) Since \(3\left(x+\frac52\right)^2\geq0\), the minimum value is

\(-\frac{155}{4},\)

and it occurs when

\(x=-\frac52.\)

(c) Let

\(u=y^{1/3}.\)

Then \(y^{2/3}=u^2\), so the equation becomes

\(3u^2+15u-20=0.\)

Using the completed square form,

\(3\left(u+\frac52\right)^2-\frac{155}{4}=0.\)

Thus

\(3\left(u+\frac52\right)^2=\frac{155}{4},\)

so

\(u=-\frac52\pm\sqrt{\frac{155}{12}}.\)

Since \(u=y^{1/3}\),

Therefore

\(y=1.31\quad\text{or}\quad y=-226\)

to 3 significant figures.

Answer: \((x-3)^2-8\), minimum point \((3,-8)\).

(a) Complete the square:

\(x^2-6x+1=(x-3)^2-9+1.\)

Therefore

\(x^2-6x+1=(x-3)^2-8.\)

(b) Since \((x-3)^2\geq0\), the minimum value of \(y\) is \(-8\), which occurs when \(x=3\).

So the minimum point is

\((3,-8).\)

Answer: \(9\left(x-\dfrac23\right)^2+1\); minimum point \(\left(\dfrac23,1\right)\).

(a) Complete the square:

\(9x^2-12x+5 =9\left(x^2-\frac43x\right)+5.\)

Now

\(x^2-\frac43x =\left(x-\frac23\right)^2-\frac49.\)

Therefore

\(9x^2-12x+5 =9\left[\left(x-\frac23\right)^2-\frac49\right]+5.\)

So

\(9x^2-12x+5 =9\left(x-\frac23\right)^2-4+5.\)

Hence

\(9x^2-12x+5=9\left(x-\frac23\right)^2+1.\)

(b) From the completed-square form, the minimum point is

\(\left(\frac23,1\right).\)

Answer: \(5\left(x-\dfrac32\right)^2-\dfrac{41}{4}\); least value \(-\dfrac{41}{20}\) at \(x=\dfrac32\).

Complete the square:

\(5x^2-15x+1=5\left(x^2-3x\right)+1.\)

Now

\(x^2-3x=\left(x-\frac32\right)^2-\frac94.\)

Therefore

\(5x^2-15x+1 =5\left[\left(x-\frac32\right)^2-\frac94\right]+1.\)

So

\(5x^2-15x+1 =5\left(x-\frac32\right)^2-\frac{45}{4}+1 =5\left(x-\frac32\right)^2-\frac{41}{4}.\)

Since

\(x^2-3x+0.2=\frac15(5x^2-15x+1),\)

we have

\(x^2-3x+0.2=\left(x-\frac32\right)^2-\frac{41}{20}.\)

The square term is always non-negative, so the least value is

\(\boxed{-\frac{41}{20}}\)

and it occurs when

\(\boxed{x=\frac32}.\)

Answer: \(y=2(x-1)^2-9\); minimum \(-9\) when \(x=1\); \(p=9.74\).

Complete the square:

\(y=2x^2-4x-7=2(x^2-2x)-7.\)

Since \(x^2-2x=(x-1)^2-1\),

\(y=2\bigl((x-1)^2-1\bigr)-7 =2(x-1)^2-9.\)

So the minimum value is \(-9\), occurring when

\(x=1.\)

For \(2p-4\sqrt p-7=0\), let

\(x=\sqrt p.\)

Then

\(2x^2-4x-7=0.\)

Using the completed square form,

\(2(x-1)^2-9=0.\)

Hence

\((x-1)^2=\frac92.\)

Since \(x=\sqrt p\geq0\),

\(x=1+\frac3{\sqrt2}.\)

Therefore

\(p=x^2=\left(1+\frac3{\sqrt2}\right)^2=9.7426\ldots.\)

So

\(\boxed{p=9.74}.\)

Answer: (i) \(\dfrac{81}{4}-\left(x-\dfrac72\right)^2\). (ii) Maximum \(=\dfrac{81}{4}\) at \(x=\dfrac72\). (iii) \(z=\pm\sqrt8\).

(i) Complete the square:

\(8+7x-x^2=-(x^2-7x-8).\)

Now

\(x^2-7x=\left(x-\frac72\right)^2-\frac{49}{4}.\)

So

\(8+7x-x^2 = -\left[\left(x-\frac72\right)^2-\frac{49}{4}-8\right].\)

Since \(8=\dfrac{32}{4}\), this becomes

\(8+7x-x^2 = \frac{81}{4}-\left(x-\frac72\right)^2.\)

(ii) The square term is always non-negative, so the expression is largest when

\(\left(x-\frac72\right)^2=0.\)

Therefore the maximum value is

\(\frac{81}{4}\)

and it occurs when

\(x=\frac72.\)

(iii) Put \(x=z^2\) into the completed-square form:

\(8+7z^2-z^4 = \frac{81}{4}-\left(z^2-\frac72\right)^2.\)

Set this equal to zero:

\(\left(z^2-\frac72\right)^2=\frac{81}{4}.\)

Taking square roots gives

\(z^2-\frac72=\pm\frac92.\)

So

\(z^2=8\quad\text{or}\quad z^2=-1.\)

There are no real solutions from \(z^2=-1\). Hence

\(\boxed{z=\pm\sqrt8}.\)