Anna P.

asked • 12d# Write an equation for the quadratic function in vertex form and standard form. (Make sure to find the value of a.)

There is a picture of a graph with the vertex (15,25). A point on the graph would be (10,12.5).

## 3 Answers By Expert Tutors

Chiara L. answered • 12d

HS/MS Math Teacher for Integrated Math and Tradition Math

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

plug in the vertex (h, k) and the point into (x,y) in the above equation and solve for a.

12.5=a(10-15)^{2}+25

12.5=25a+25

-12.5=25a

-0.5=a

Plug a and the vertex into vertex form

y=-0.5(x-15)^{2}+25

Multiply vertex form out to get standard form

y=-0.5(x-15)(x-15)+25

y=-0.5(x^{2}-30x+225)+25

y=-0.5x^{2}+15x-112.5+25

y=-0.5x^{2}+15x-87.5

answer:

y=-0.5(x-15)^{2}+25

y=-0.5x^{2}+15x-87.5

Mark M. answered • 12d

Mathematics Teacher - NCLB Highly Qualified

f(x) = a(x - 15)^{2} + 25

12.5 = a(10 - 15)^{2} + 25

Can you solve for a and answer?

William W. answered • 12d

Math and science made easy - learn from a retired engineer

The vertex form of a quadratic is f(x) = a(x - h)^{2} + k where the vertex is (h, k) so plug in the vertex of (15, 25) to get: f(x) = a(x - 15)^{2} + 25

To find a, plug in the point (10, 12,5)

So 12.5 = a(10 - 15)^{2} + 25 and solve

12.5 = 25a + 25

-12.5 = 25a

a = -1/2

So the equation is f(x) = -1/2(x - 15)^{2} + 25

To get it in standard form, multiply it out and combine like-term start by getting (x - 15)^{2} which is x^{2} - 30x + 225, multiply that by -1/2 then add 25

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Anna P.

Thank you!!12d