Question

graphing a quadratic function in vertex form
graph ( h(x) = 0.5(x + 2)^2 - 4 ) by following these steps:

step 1: identify ( a ), ( h ), and ( k ).
( a = 0.5 ) ( h = -2 ) ( k = -4 )

step 2: plot the vertex at ( (-2, -4) ).

step 3: the axis of symmetry is the line
( x = -2 )

step 4: evaluate the function at two other ( x )-values:
( h(-4) = -2 ) ( h(-6) = )

(options for ( h(-6) ): -2, 2, 4)

Explanation:

Step1: Recall vertex form definition

The vertex form of a quadratic function is $h(x)=a(x-h)^2+k$. For $h(x)=0.5(x+2)^2-4$, rewrite as $h(x)=0.5(x-(-2))^2+(-4)$, so $a=0.5$, $h=-2$, $k=-4$.

Step2: Calculate $h(-6)$

Substitute $x=-6$ into the function:

$$\begin{align*} h(-6)&=0.5(-6+2)^2-4\\ &=0.5(-4)^2-4\\ &=0.5(16)-4\\ &=8-4 \end{align*}$$

Answer:

$h(-6)=4$

Additionally, completing the table:

$x$	$y$
$-4$	$-2$
$-2$	$-4$
$0$	$-2$
$2$	$1$

The axis of symmetry is $x=-2$, and the vertex is $(-2, -4)$.