Integrate the gradient:

\(y=\int(3x^2-6x+2)\,dx=x^3-3x^2+2x+c\).

Now use the point \((-2,-10)\):

\(-10=(-2)^3-3(-2)^2+2(-2)+c\)

\(-10=-8-12-4+c\)

\(-10=-24+c\)

\(c=14\).

So the equation is \(y=x^3-3x^2+2x+14\).

First expand:

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

Integrate:

\(y=\int(1-4x+4x^2)\,dx=x-2x^2+\frac{4}{3}x^3+c\).

Use the point \((1,8)\):

\(8=1-2+\frac{4}{3}+c\)

\(8=\frac{1}{3}+c\)

\(c=\frac{23}{3}\).

So \(y=x-2x^2+\frac{4}{3}x^3+\frac{23}{3}\).

Expand first:

\(x(2x+5)=2x^2+5x\).

Integrate:

\(y=\int(2x^2+5x)\,dx=\frac{2}{3}x^3+\frac{5}{2}x^2+c\).

Use \((5,-1)\):

\(-1=\frac{2}{3}(5)^3+\frac{5}{2}(5)^2+c\)

\(-1=\frac{250}{3}+\frac{125}{2}+c\)

\(-1=\frac{875}{6}+c\)

\(c=-\frac{881}{6}\).

So the equation is \(y=\frac{2}{3}x^3+\frac{5}{2}x^2-\frac{881}{6}\).

Simplify the gradient:

\(\sqrt{x}(\sqrt{x}-3)=x-3x^{1/2}\).

Integrate:

\(y=\int(x-3x^{1/2})\,dx=\frac{1}{2}x^2-2x^{3/2}+c\).

Use \((9,12)\):

\(12=\frac{1}{2}(9)^2-2(9)^{3/2}+c\)

\(12=\frac{81}{2}-54+c\)

\(12=-\frac{27}{2}+c\)

\(c=\frac{51}{2}\).

So the equation is \(y=\frac{1}{2}x^2-2x^{3/2}+\frac{51}{2}\).

Simplify the gradient:

\(\dfrac{9x^3-3x}{x}=9x^2-3\).

Integrate:

\(y=\int(9x^2-3)\,dx=3x^3-3x+c\).

Use \((-5,4)\):

\(4=3(-5)^3-3(-5)+c\)

\(4=-375+15+c\)

\(4=-360+c\)

\(c=364\).

So the equation is \(y=3x^3-3x+364\).

Expand first:

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

Integrate:

\(y=\int(15x^2+x-2)\,dx=5x^3+\frac{1}{2}x^2-2x+c\).

Use \((-4,-6)\):

\(-6=5(-4)^3+\frac{1}{2}(-4)^2-2(-4)+c\)

\(-6=-320+8+8+c\)

\(-6=-304+c\)

\(c=298\).

So the equation is \(y=5x^3+\frac{1}{2}x^2-2x+298\).

Rewrite using powers:

\(\dfrac{5}{\sqrt{x}}-10\sqrt{x^3}=5x^{-1/2}-10x^{3/2}\).

Integrate:

\(y=\int(5x^{-1/2}-10x^{3/2})\,dx=10x^{1/2}-4x^{5/2}+c\).

Use \((1,-6)\):

\(-6=10(1)^{1/2}-4(1)^{5/2}+c\)

\(-6=6+c\)

\(c=-12\).

So the equation is \(y=10x^{1/2}-4x^{5/2}-12\).

Rewrite the gradient as \(6x^{-3}+2\).

Integrate:

\(y=\int(6x^{-3}+2)\,dx=-3x^{-2}+2x+c\).

So \(y=-\dfrac{3}{x^2}+2x+c\).

Use \((7,10)\):

\(10=-\dfrac{3}{49}+14+c\)

\(10=\dfrac{683}{49}+c\)

\(c=-\dfrac{193}{49}\).

Hence \(y=-\dfrac{3}{x^2}+2x-\dfrac{193}{49}\).

Expand:

\(f\,\!'(x)=x(3-x)=3x-x^2\).

Integrate:

\(f(x)=\int(3x-x^2)\,dx=\frac{3}{2}x^2-\frac{1}{3}x^3+c\).

Use \((-2,-1)\):

\(-1=\frac{3}{2}(-2)^2-\frac{1}{3}(-2)^3+c\)

\(-1=6+\frac{8}{3}+c\)

\(-1=\frac{26}{3}+c\)

\(c=-\frac{29}{3}\).

So the equation is \(y=\frac{3}{2}x^2-\frac{1}{3}x^3-\frac{29}{3}\).

Integrate the second derivative:

\(\frac{dy}{dx}=\int -8x\,dx=-4x^2+c\).

At a maximum point, the gradient is 0.

So when \(x=1\),

\(0=-4(1)^2+c\)

\(c=4\).

Hence \(\frac{dy}{dx}=4-4x^2\).

Integrate again:

\(y=\int(4-4x^2)\,dx=4x-\frac{4}{3}x^3+k\).

Use \((2,-1)\):

\(-1=4(2)-\frac{4}{3}(2)^3+k\)

\(-1=8-\frac{32}{3}+k\)

\(-1=-\frac{8}{3}+k\)

\(k=\frac{5}{3}\).

So the equation is \(y=4x-\frac{4}{3}x^3+\frac{5}{3}\).

Integrate:

\(f(x)=\int(8x^3-4+3x^{-1/2})\,dx\)

\(f(x)=2x^4-4x+6x^{1/2}+c\).

Use \(f(4)=3\):

\(3=2(4)^4-4(4)+6(4)^{1/2}+c\)

\(3=512-16+12+c\)

\(3=508+c\)

\(c=-505\).

So \(f(x)=2x^4-4x+6x^{1/2}-505\).

Integrate once:

\(\frac{dy}{dx}=\int(-3x+2)\,dx=-\frac{3}{2}x^2+2x+c\).

Use \(x=-1\), \(\frac{dy}{dx}=5\):

\(5=-\frac{3}{2}(-1)^2+2(-1)+c\)

\(5=-\frac{3}{2}-2+c\)

\(5=-\frac{7}{2}+c\)

\(c=\frac{17}{2}\).

So \(\frac{dy}{dx}=-\frac{3}{2}x^2+2x+\frac{17}{2}\).

Integrate again:

\(y=-\frac{1}{2}x^3+x^2+\frac{17}{2}x+k\).

Use \(x=-1\), \(y=0\):

\(0=-\frac{1}{2}(-1)^3+(-1)^2+\frac{17}{2}(-1)+k\)

\(0=\frac{1}{2}+1-\frac{17}{2}+k\)

\(0=-7+k\)

\(k=7\).

Hence \(y=-\frac{1}{2}x^3+x^2+\frac{17}{2}x+7\).

Integrate:

\(y=\int(6x^2-4x+3)\,dx=2x^3-2x^2+3x+c\).

Use \((3,10)\):

\(10=2(3)^3-2(3)^2+3(3)+c\)

\(10=54-18+9+c\)

\(10=45+c\)

\(c=-35\).

So the equation is \(y=2x^3-2x^2+3x-35\).