Vertex and Zeros of a Quadratic Function Practice

To find the zeros of the function, we need to solve the equation $f(x)=0$. We can do that by factoring $f(x)$.

$\begin{aligned}x^2-10x-56&=0\\\\(x-14)(x+4)&=0\\\\x-14=0&\text{ or }x+4=0\\\\x= {14}&\text{ or }x= {-4}\end{aligned}$

There are many ways to find the vertex. We will do it by using the fact that the $x$-coordinate of the vertex is exactly between the two zeros.

$\begin{aligned}\text{vertex's }x\text{-coordinate}&=\dfrac{( {14})+( {-4})}{2}\\\\&= {5}\end{aligned}$

Now we can find the vertex's $y$-coordinate by evaluating $f( {5})$:

$\begin{aligned}f( {5})&=( {5})^2-10( {5})-56\\\\&=25-50-56\\\\&=-81\end{aligned}$

In conclusion,

$\begin{aligned}\text{smaller }x&=-4\\\\\text{larger }x&=14\end{aligned}$

The vertex of the parabola is at $(5,-81)$

$\begin{aligned}(r-14)^2-56.25&=0\\\\(r-14)^2&=56.25\\\\\sqrt{(r-14)^2}&=\sqrt{56.25}\\\\r-14&=\pm 7.5\\\\r&=\pm7.5+14\\\\r= {6.5}&\text{ or }r= {21.5}\end{aligned}$

$f(r)$ is given in vertex form:

$f(r)=(r- {14})^2 {-56.25}$

So the vertex of the parabola is at $( {14}, {-56.25})$.

In conclusion,

$\begin{aligned}\text{smaller }r&=6.5\\\\\text{larger }r&=21.5\end{aligned}$

The vertex of the parabola is at $(14,-56.25)$

$\begin{aligned}-(r-12)(r+3)&=0\\\\r-12=0&\text{ or }r+3=0\\\\r= {12}&\text{ or }r= {-3}\end{aligned}$

There are many ways to find the vertex. We will do it by using the fact that the $r$-coordinate of the vertex is exactly between the two zeros.

$\begin{aligned}\text{vertex's }r\text{-coordinate}&=\dfrac{( {12})+( {-3})}{2}\\\\&= {\dfrac{9}{2}}\end{aligned}$

Now we can find the vertex's $y$-coordinate by evaluating $f\left( {\dfrac{9}{2}}\right)$

$\begin{aligned}f\left( {\dfrac{9}{2}}\right)&=-\left( {\dfrac{9}{2}}-12\right)\left( {\dfrac{9}{2}}+3\right)\\\\&=-\left(-\dfrac{15}{2}\right)\left(\dfrac{15}{2}\right)\\\\&=\dfrac{225}{4}\end{aligned}$

In conclusion,

$\begin{aligned}\text{smaller }r&=-3\\\\\text{larger }r&=12\end{aligned}$

The vertex of the parabola is at $\left(\dfrac{9}{2},\dfrac{225}{4}\right)$

To find the zeros of the function, we need to solve the equation $h(x)=0$. We can do that by factoring $h(x)$.

$\begin{aligned}x^2-7x-8&=0\\\\(x-8)(x+1)&=0\\\\x-8=0&\text{ or }x+1=0\\\\x= {8}&\text{ or }x= {-1}\end{aligned}$

There are many ways to find the vertex. We will do it by using the fact that the $x$-coordinate of the vertex is exactly between the two zeros.

$\begin{aligned}\text{vertex's }x\text{-coordinate}&=\dfrac{( {8})+( {-1})}{2}\\\\&= {\dfrac{7}2}\end{aligned}$

Now we can find the vertex's $y$-coordinate by evaluating $h\left( {\dfrac{7}2}\right)$:

$\begin{aligned}h\left( {\dfrac{7}2}\right)&=\left( {\dfrac{7}2}\right)^2-7\left( {\dfrac{7}2}\right)-8\\\\&=\dfrac{49}{4}-\dfrac{49}{2}-8\\\\&=-\dfrac{81}{4}\end{aligned}$

In conclusion,

$\begin{aligned}\text{smaller }x&=-1\\\\\text{larger }x&=8\end{aligned}$

The vertex of the parabola is at $\left(\dfrac{7}2,-\dfrac{81}{4}\right)$