Chapter SH, Problem 9.4EP

To determine

To find: The vertex, the equation of the axis of symmetry, and the y -intercept of the graph of the equation $y=4x^2+16x-6$ .

Answer to Problem 9.4EP

The vertex, the equation of the axis of symmetry, and the y -intercept of the graph of the equation $y=4x^2+16x-6$ are $\left(-2,-22\right)$ , $x=-2$ and $-6$ , respectively.

Explanation of Solution

Given information:

The given equation is $y=4x^2+16x-6$ .

Calculation:

The given equation is a quadratic equation.

  $y=a{x}^2+bx+c$

Compare the given equation with the standard form of the quadratic equation.

  $a=4,b=16,c=-6$

Write the equation for the axis of the symmetry.

  $x=\frac{-b}{2a}$

Substitute $16$ for $b$ and 4 for $a$ .

  $\begin{array}{c}x=\frac{-16}{2\left(4\right)}\\ =\frac{-16}{8}\\ =-2\end{array}$

So, the equation for the axis of the symmetry is $x=-2$ .

The vertex is equal to $\left(x,y\right)$ . To calculate the value of $y$ , substitute $-2$ for $x$ in the given equation.

  $\begin{array}{c}y=4{\left(-2\right)}^2+16\times \left(-2\right)-6\\ =16-32-6\\ =10-32\\ =-22\end{array}$

So, the vertex is $\left(-2,-22\right)$ .

The y -intercept of the graph of the equation is equal to the value of $c$ . So, the y -intercept of the graph of the equation is $-6$ .

Therefore, the vertex, the equation of the axis of symmetry, and the y -intercept of the graph of the equation $y=4x^2+16x-6$ are $\left(-2,-22\right)$ , $x=-2$ and $-6$ , respectively.