Chapter 9.5, Problem 12CYU

To determine

Find the vertex form

Answer to Problem 12CYU

  $y=3{\left(x+2\right)}^2-51$

Explanation of Solution

Given Information: $y=3x^2+12x-39$

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).

  $y=3x^2+12x-39$

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 $y=3x^2+12x-39$ with $a{x}^2+bx+c$

  $a=3,b=12$ and $c=-39$

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

As,

  $h=\frac{-b}{2a}=\frac{-\left(12\right)}{2\times 3}=\frac{-12}{6}$

Hence,

  $h=-2$

Now, to find $k$ vertex, put value of $h$ vertex in $y=3x^2+12x-39$

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

  $k=3\times {\left(-2\right)}^2+12\times \left(-2\right)-39$

So, simplify

  $k=12-24-39$

Hence,

  $k=-51$

Now,

Vertex $\left(h,k\right)=\left(-2,-51\right)$

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

As, $a=3$ and Vertex $\left(h,k\right)=\left(-2,-51\right)$

  $y=3{\left(x-\left(-2\right)\right)}^2-51$

Hence,

  $y=3{\left(x+2\right)}^2-51$