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Home Q & A Board Science-Related Homework Help Mathematics
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huntertrust huntertrust
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11 years ago
Write an equation of a quadratic function whose graph is a parabola that has a vertex (-3,7) and that passes through the origin.

How? Please help!!

Many thanks!
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leogetsleogets
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#1
11 years ago
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wrote...
#2
11 years ago
y = x^2+6x+16
wrote...
#3
11 years ago
We have one point, the root (0 , 0) and vertex at (h , k) ---> (-3 , 7)

The parabola in vertex form is:

y = a(x - h)^2 + k

We are given the vertex :

y =a (x +3 )^2 + 7

Now we can sub in the root's coordinates and search for a:

0 =a (0 + 3)^2 + 7

0 =9a + 7

-7 = 9a

a =-7/9

We now have the vertex form of the equation:

y = (-7/9)(x+3)^2 + 7

You can expand and simplify to put in general or standard form.

y = (-7/9)x^2 - 14x / 3
wrote...
#4
11 years ago
a quadratic has the form of the following:

 y = a(x-h)^2 + k

the vertex is (h,k)

therefore, it becomes the following.

y = a(x+3)^2 + 7

the fact that (0,0) is a point on the graph makes the equation as the folowing.

0 = a(0+3)^2 + 7

0 = 9a + 7

-7/9 = a

so the equation is the following:

y = (-7/9)(x + 3)^2 + 7
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