Chapter 9.5, Problem 10CYU

To determine

Find the vertex form

Answer to Problem 10CYU

  $y={\left(x-6\right)}^2-20$

Explanation of Solution

Given Information: $y=x^2-12x+16$

Formula Used: Vertex form $y=a{\left(x-h\right)}^2+k$ , $h=\frac{-b}{2a}$ and $k=f\left(h\right)$

Calculation:

When written in "vertex form":

•  $\left(h,k\right)$  is the vertex of the parabola, and  $x=h$ and $y=k$ is the axis of symmetry.

• $h$  represents a horizontal shift (how far left, or right, the graph has shifted from  x  = 0).

• $k$  represents a vertical shift (how far up, or down, the graph has shifted from  y  = 0).

  $y=x^2-12x+16$

A quadratic equation is written as  $a{x}^2+bx+c$  in its standard form.

The vertex $\left(h,k\right)$ can be found by using the formula  $h=\frac{-b}{2a}$ and $k=f\left(h\right)$

Now, to find $h$ vertex, compare equation $y=x^2-12x+16$ with $a{x}^2+bx+c$

  $a=1,b=-12$ and $c=16$

Now, put value of a, b and c in $h=\frac{-b}{2a}$

As,

  $h=\frac{-b}{2a}=\frac{-\left(-12\right)}{2\times 1}=\frac{12}{2}$

Hence,

  $h=6$

Now, to find $k$ vertex, put value of $h$ vertex in $y=x^2-12x+16$

As $x=h$ and $y=k$

  $k=6^2-\left(12\times 6\right)+16$

So, simplify

  $k=36-72+16$

Hence,

  $k=-20$

Now,

Vertex $\left(h,k\right)=\left(6,-20\right)$

As, Vertex form $y=a{\left(x-h\right)}^2+k$

As, $a=1$ and Vertex $\left(h,k\right)=\left(6,-20\right)$

  $y=1{\left(x-6\right)}^2-20$

Hence,

  $y={\left(x-6\right)}^2-20$