Answer: D. y = -(x+3)^2
======================================================
Explanation:
Recall that vertex form is
y = a(x-h)^2 + k
where (h,k) is the vertex. We want the vertex to be (-3,0) which means that h = -3 and k = 0 must be the case.
Plug those values in to get
y = a(x-h)^2 + k
y = a(x-(-3))^2 + 0
y = a(x+3)^2
Now we need the value of 'a', which determines two things
- How vertically stretched or compressed the graph is, and,
- If the parabola opens upward or downward. We have a > 0 for upward and a < 0 for downward.
To determine 'a', we'll use the other point (-5,-4).
So we'll plug in x = -5 and y = -4 and solve for 'a'
y = a(x+3)^2
-4 = a(-5+3)^2
-4 = a(-2)^2
-4 = a(4)
-4 = 4a
4a = -4
a = -4/4
a = -1
The equation y = a(x+3)^2 updates to y = -1(x+3)^2 which simplifies to the final answer y = -(x+3)^2. The answer is choice D.
---------------
As a check, plugging x = -5 should lead to y = -4
y = -(x+3)^2
y = -(-5+3)^2
y = -(-2)^2
y = -(4)
y = -4
This helps confirm the answer.
I'll let you check to see if (-3,0) is on this graph. Plugging in x = -3 should lead to y = 0.
