Question

a. (1,1)
b. (0,-2)
c. (3,1)
d. (2,2)

Explanation:

Step1: Identify parabola vertex

The vertex of the parabola is at $(2,2)$. The standard form of a downward-opening parabola is $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex. Substituting $h=2, k=2$, we get $y = a(x-2)^2 + 2$.

Step2: Solve for coefficient $a$

Use the y-intercept $(0,-2)$ from the graph. Substitute $x=0, y=-2$ into the equation:
$-2 = a(0-2)^2 + 2$
$-2 = 4a + 2$
$4a = -4$
$a = -1$
So the equation is $y = -(x-2)^2 + 2$.

Step3: Test each option

• Option A: Substitute $x=1$: $y = -(1-2)^2 + 2 = -1 + 2 = 1$. The point $(1,1)$ is on the parabola, but check other options.
• Option B: Substitute $x=0$: $y = -(0-2)^2 + 2 = -4 + 2 = -2$. The point $(0,-2)$ is on the parabola.
• Option C: Substitute $x=3$: $y = -(3-2)^2 + 2 = -1 + 2 = 1$. The point $(3,1)$ is on the parabola.
• Option D: Substitute $x=2$: $y = -(2-2)^2 + 2 = 0 + 2 = 2$. The point $(2,2)$ is the vertex, on the parabola.

(Note: Assuming the question asks for the vertex, the unique highest point of the parabola, which is $(2,2)$.)

Answer:

D. (2,2)