# pre cal

how do you determine the the quadratic function if the problem gives you a graph with the line has points of (-3,5) and (0,-4)?

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asked by shelly
1. Two points only is not enough to establish a quadratic, unless one of those points is the vertex.

If one of the points is a vertex, use the method I just showed you in your previous post.
Let me know what you think.

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posted by Reiny
2. the (-3,5) is the vertex how do i use that i am not completely understand

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posted by shelly
3. where do i plug in thoes number

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posted by shelly
4. Look back at my explanation in the previous post
for vertex (h,k), we get
y = a(x-h)^2 + k

so for vertex (-3,5), we get

y = a(x-(-3) )^2 + 5
or
y = a(x + 3)^2 + 5

Now plug in the other point (0, -4)
-4 = a(0+3)^2 + 5
-4 = a(9) + 5
-9 = 9a
a = -1

y = -(x+3)^2 + 5

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posted by Reiny
5. how did you get the nine after the A?

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posted by shelly
6. mmmhhh?

(0+3)^2
= 3^2
= 9

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posted by Reiny

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