Chapter 1.2, Problem 4E

Solution

Graph the function. Label the vertex and axis of symmetry.

Axis of symmetry (x) $=\frac{-b}{2a}=-4$

Vertex $=(-4,0)$

Given:

The given function $y={(x+4)}^2$

Concept Used:

• The vertex is the highest point if the parabola opens downward and the lowest point if the parabola opens upward.
• The axis of symmetry is the line that cuts the parabola into 2 matching halves and the vertex lies on the axis of symmetry.

Calculation:

The graph of the given function $y={(x+4)}^2$ is -

  

The given function $y={(x+4)}^2$

  $\begin{array}{c}$ y={(x+4)}^2$y=x^2+16+2\cdot x\cdot 4\\ y=x^2+16+8x\\ y=x^2+8x+16\end{array}$

Here $a=1,b=8,c=16$

Axis of symmetry (x) $=\frac{-b}{2a}$

  $\begin{array}{l}=\frac{-(8)}{2(1)}\\ =\frac{-8}{2}\\ =-4\end{array}$

So Axis of symmetry (x) $=\frac{-b}{2a}=-4$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=y\left(-4\right)\\ $ y={(x+4)}^2$$ y={(-4+4)}^2$=0\end{array}$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

Vertex $=(-4,0)$