Practice questions

The differences are ${5-(-3)=8}$ and ${-1-5=-6}$. Use the distance formula:

$\begin{align*}&d\\&=\sqrt{8^2+(-6)^2}\\&=\sqrt{64+36}=\sqrt{100}=10\end{align*}$

The midpoint is the average of the coordinates:

$\left(\dfrac{-3+5}{2},\dfrac{5+(-1)}{2}\right)=(1,2)$

Group the ${x}$-terms and the ${y}$-terms, and move ${-12}$ to the right side:

${(x^2+4x)+(y^2-6y)=12}$

Complete both squares. Add ${\left(\dfrac{4}{2}\right)^2=4}$ and ${\left(\dfrac{-6}{2}\right)^2=9}$ to both sides:

$\begin{align*}(x^2+4x+4)+(y^2-6y+9)&=12+4+9\\(x+2)^2+(y-3)^2&=25\end{align*}$

The center is ${(-2,3)}$, and the radius is ${\sqrt{25}=5}$.

For the ${y}$-intercept, put ${x=0}$: ${y=-8}$.

For the ${x}$-intercepts, put ${y=0}$, and factor:

$\begin{align*}x^2-2x-8&=0\\(x-4)(x+2)&=0\end{align*}$

So ${x=4}$ or ${x=-2}$. The intercepts are ${(0,-8)}$, ${(4,0)}$, and ${(-2,0)}$.

${x}$-axis: Replace ${y}$ by ${-y}$. This gives ${-y=x^4+3}$, which is a different equation. No symmetry.

${y}$-axis: Replace ${x}$ by ${-x}$. Since ${(-x)^4=x^4}$, this gives ${y=x^4+3}$ again. The graph is symmetric about the ${y}$-axis.

Origin: Replace both. This gives ${-y=x^4+3}$, a different equation. No symmetry.

The slope is

$m=\dfrac{-8-1}{4-(-2)}=\dfrac{-9}{6}=-\dfrac{3}{2}$.

Use point-slope form with ${(-2,1)}$:

$\begin{align*}y-1&=-\dfrac{3}{2}(x+2)\\y-1&=-\dfrac{3}{2}x-3\\y&=-\dfrac{3}{2}x-2\end{align*}$

Check: At ${x=4}$, ${y=-6-2=-8}$. It checks.

Solve ${2x+5y=10}$ for ${y}$ to find its slope:

${y=-\dfrac{2}{5}x+2}$

A parallel line has the same slope, ${-\dfrac{2}{5}}$. Use the point ${(1,-3)}$:

$\begin{align*}y+3&=-\dfrac{2}{5}(x-1)\\y&=-\dfrac{2}{5}x+\dfrac{2}{5}-3\\y&=-\dfrac{2}{5}x-\dfrac{13}{5}\end{align*}$

The given line has slope ${3}$. A perpendicular line has the negative reciprocal slope, ${-\dfrac{1}{3}}$. Use the point ${(6,2)}$:

$\begin{align*}y-2&=-\dfrac{1}{3}(x-6)\\y-2&=-\dfrac{1}{3}x+2\\y&=-\dfrac{1}{3}x+4\end{align*}$

$\begin{align*}&f(-2)\\&=2(-2)^2-(-2)+3\\&=8+2+3=13\end{align*}$

For ${f(a-1)}$, replace every ${x}$ by ${(a-1)}$:

$\begin{align*}&f(a-1)\\&=2(a-1)^2-(a-1)+3\\&=2(a^2-2a+1)-a+1+3\\&=2a^2-4a+2-a+4\\&=2a^2-5a+6\end{align*}$

First find ${f(x+h)}$:

$\begin{align*}&f(x+h)\\&=(x+h)^2+4(x+h)\\&=x^2+2xh+h^2+4x+4h\end{align*}$

Subtract ${f(x)=x^2+4x}$. The terms ${x^2}$ and ${4x}$ cancel:

${f(x+h)-f(x)=2xh+h^2+4h}$

Divide by ${h}$:

${\dfrac{2xh+h^2+4h}{h}=2x+h+4}$

The number under the square root cannot be negative:

$\begin{align*}6-2x&\ge 0\\-2x&\ge -6\\x&\le 3\end{align*}$

(Dividing by ${-2}$ reverses the inequality sign.) The domain is ${(-\infty,3]}$.

The denominator cannot be ${0}$. Factor it:

${x^2-5x+6=(x-2)(x-3)}$

It is ${0}$ at ${x=2}$ and ${x=3}$. The domain is all real numbers except ${2}$ and ${3}$:

${(-\infty,2)\cup(2,3)\cup(3,\infty)}$

Odd powers change sign when ${x}$ becomes ${-x}$:

$\begin{align*}&f(-x)\\&=(-x)^5-2(-x)^3\\&=-x^5+2x^3\\&=-(x^5-2x^3)=-f(x)\end{align*}$

So ${f}$ is odd. For ${g}$, both ${(-x)^2=x^2}$ and ${|-x|=|x|}$:

${g(-x)=x^2+|x|=g(x)}$

So ${g}$ is even.

The ${+3}$ inside the root moves the graph ${3}$ to the left. The ${-1}$ outside moves it ${1}$ down. So the starting point ${(0,0)}$ moves to ${(-3,-1)}$.

The graph is the V of ${y=|x|}$ moved ${2}$ left. Its corner is at ${(-2,0)}$. The graph falls to the left of the corner and rises to the right of it. So ${f}$ is decreasing on ${(-\infty,-2)}$ and increasing on ${(-2,\infty)}$.

Half of ${8}$ is ${4}$, and ${4^2=16}$. Add and subtract ${16}$:

$\begin{align*}&f(x)\\&=x^2+8x+16-16+10\\&=(x+4)^2-6\end{align*}$

The vertex is ${(-4,-6)}$.

Here ${a=-3}$ and ${b=12}$. The vertex is at

${x=-\dfrac{12}{2(-3)}=2}$.

Its ${y}$-value is

${f(2)=-12+24-5=7}$.

The vertex is ${(2,7)}$. Since ${a<0}$, the parabola opens downward. So the maximum value is ${7}$, and the range is ${(-\infty,7]}$.

Call one number ${x}$. Then the other is ${30-x}$. Their product is

${P(x)=x(30-x)=-x^2+30x}$.

Here ${a=-1<0}$, so the vertex gives the maximum. It is at

${x=-\dfrac{30}{2(-1)}=15}$.

The other number is also ${15}$. The largest product is ${15\cdot 15=225}$.

The ball is highest at the vertex, because ${a=-4.9<0}$. The vertex is at

$t=-\dfrac{24.5}{2(-4.9)}=\dfrac{24.5}{9.8}=2.5$.

The height then is

$\begin{align*}&h(2.5)\\&=-4.9(2.5)^2+24.5(2.5)+1\\&=-30.625+61.25+1\\&=31.625\end{align*}$

The ball is highest after ${2.5}$ seconds, at about ${31.6}$ m.

Add the two formulas:

$\begin{align*}&(f+g)(x)\\&=3x-2+x^2+x\\&=x^2+4x-2\end{align*}$

For the product, ${f(1)=1}$ and ${g(1)=2}$. So ${(fg)(1)=1\cdot 2=2}$.

Put ${g(x)}$ into ${f}$:

$\begin{align*}&(f\circ g)(x)\\&=3(x^2+x)-2\\&=3x^2+3x-2\end{align*}$

Put ${f(x)}$ into ${g}$:

$\begin{align*}&(g\circ f)(x)\\&=(3x-2)^2+(3x-2)\\&=9x^2-12x+4+3x-2\\&=9x^2-9x+2\end{align*}$

Work from the inside out. First ${g(-2)=4-2=2}$. Then ${f(2)=6-2=4}$. So ${(f\circ g)(-2)=4}$.

The formula is ${F(G(x))=\sqrt{1-x^2}}$. Every ${x}$ can go into ${G}$. But ${F}$ needs an input that is not negative:

$\begin{align*}1-x^2&\ge 0\\x^2&\le 1\end{align*}$

This holds when ${-1\le x\le 1}$. The domain is ${[-1,1]}$.

The inside part is ${x^2-3}$, and it is raised to the fourth power. So take ${q(x)=x^2-3}$ and ${p(x)=x^4}$.

Check: ${p(q(x))=(x^2-3)^4=h(x)}$.