Chapter 4, Problem 22PT

To determine

To find: The quadratic function in vertex form.

Answer to Problem 22PT

.The quadratic function in vertex form is $a=3$ .

Explanation of Solution

Given information:

The given equation is $3x^2+6x=2+y$ .

Calculation:

Calculate the equation.

  $\begin{array}{c}3x^2+6x=2+y\\ y=3x^2+6x-2\end{array}$

Add and subtract $3$ in the equation.

  $\begin{array}{l}y=3x^2+6x-2\\ y=3{\left(x+1\right)}^2-5\\ y=a{\left(x-h\right)}^2+k\end{array}$

Here, $a=3$ , $h=-1$ , $k=-5$ . Thus the vertex is $\left(-1,-5\right)$ . The axis of symmetry is $x=-1$ .

Plot the graph by use $Ti-83$ calculator.

  

Trace and then graph to get the below output:

  

The direction of opening of the equation is upwards. This is so because value of $a=3$ is positive.

Therefore, The quadratic function in vertex form is $\left(a=3\right)$ .