Question

graph $h(t) = 2(t - 2)^2 + 4$.

Explanation:

Step1: Identify vertex form

The function is in vertex form $h(t)=a(t-h)^2+k$, where $(h,k)$ is the vertex, $a=2$, $h=2$, $k=4$.
Vertex: $(2, 4)$

Step2: Determine parabola direction

Since $a=2>0$, parabola opens upward.

Step3: Find additional points

For $t=1$:

$h(1)=2(1-2)^2+4=2(1)+4=6$ → Point: $(1,6)$

For $t=3$:

$h(3)=2(3-2)^2+4=2(1)+4=6$ → Point: $(3,6)$

For $t=0$:

$h(0)=2(0-2)^2+4=2(4)+4=12$ → Point: $(0,12)$

For $t=4$:

$h(4)=2(4-2)^2+4=2(4)+4=12$ → Point: $(4,12)$

Answer:

Plot the vertex $(2,4)$, symmetric points $(1,6)$, $(3,6)$, $(0,12)$, $(4,12)$, then draw an upward-opening parabola passing through these points.