Question

graph the parabola.
$y=2(x-5)^2 + 3$
plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph-a-function button.

Explanation:

Step1: Identify the vertex

The vertex form of a parabola is $y=a(x-h)^2+k$, where $(h,k)$ is the vertex. For $y=2(x-5)^2+3$, $h=5$, $k=3$.
Vertex: $(5, 3)$

Step2: Find left points (x=4, x=3)

Substitute x=4:

$y=2(4-5)^2+3=2(1)+3=5$
Point: $(4, 5)$

Substitute x=3:

$y=2(3-5)^2+3=2(4)+3=11$
Point: $(3, 11)$

Step3: Find right points (x=6, x=7)

Substitute x=6:

$y=2(6-5)^2+3=2(1)+3=5$
Point: $(6, 5)$

Substitute x=7:

$y=2(7-5)^2+3=2(4)+3=11$
Point: $(7, 11)$

Answer:

Points to plot:

1. Vertex: $(5, 3)$
2. Left points: $(4, 5)$, $(3, 11)$
3. Right points: $(6, 5)$, $(7, 11)$

Connect these points to graph the upward-opening parabola.