Question

Find a and b. y= ax^2 + bx - 27; (2, -3) I can't seem to find b that's my problem...

Answer 1

is (2,-3) the vertex?

Answer 2

Yes

Answer 3

do you know vertex form?

Answer 4

If I can find what be is then I'm all set. Yeah, ax^2 to bx + C

Answer 5

To find be would be -b/2a =2?

Answer 6

I can only get to -3=a(2)^2+b(2)+C

Answer 7

vertex form is

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

we know the vertex is (2,-3), so h = 2 and k = -3

This means we get

y = a(x - 2)^2 - 3

We also know that the y-intercept is (0, -27) after plugging in x = 0 into the original equation and getting y = -27, so plug in x = 0 and y = -27 to get

-27 = a(0 - 2)^2 - 3

Solve for 'a' to get

-27 = a(0 - 2)^2 - 3

-27 = a(-2)^2 - 3

-27 = a(4) - 3

-27 = 4a - 3

4a - 3 = -27

4a = -27+3

4a = -24

a = -24/4

a = -6

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So we now know that

y = a(x - 2)^2 - 3

becomes

y = -6(x - 2)^2 - 3

To find 'b', expand this last equation out and compare it to y= ax^2 + bx - 27

Answer 8

Oh my gosh. Thank you so much! I never thought of using that equation. =) So helpful.

Answer 9

you're welcome