SOLUTION: can you tell me how to convert a parabola into an equation... x intercepts are (3,0) and (13,0) how do i find the vertex if thats all the info i have...

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: can you tell me how to convert a parabola into an equation... x intercepts are (3,0) and (13,0) how do i find the vertex if thats all the info i have...      Log On

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Question 238389: can you tell me how to convert a parabola into an equation... x intercepts are (3,0) and (13,0) how do i find the vertex if thats all the info i have...
Answer by Theo(2978) About Me  (Show Source):
You can put this solution on YOUR website!
x-intercepts are (3,0) and (13,0)

the roots of your equation are:

x = 3
x = 13

this was derived from:

x-3 = 0
x - 13 = 0

that came from:

(x-3) * (x-13) = 0

multiply these factors together to get:

x^2 - 16x + 39 = 0

that's your quadratic equation.

it's in standard form.

you can find the vertex from that form by doing the following:

standard form of a quadratic equation is:

ax^2 + bx + c = 0

your equation is:

x^2 - 16x + 39 = 0

a = 1
b = -16
c = 39

x value of the vertex is given by the equation x = -b/2a.

you get:

x = 16/2 = 8

y value of the vertex is given by the equation y = f(-b/2a).

this means y = f(8) which means you replace x with 8 in your equation and solve for y.

your equation is y = x^2 - 16x + 39

replace x with 8 to get:

y = 64 - 128 + 39 = -25.

your vertex is (x,y) = (8,-25)

a graph of your equation looks like:

graph%28400%2C400%2C-30%2C30%2C-30%2C30%2Cx%5E2+-+16x+%2B+39%29

you are really just working backwards.

you are given the equation, you find the roots.

you are given the roots, you find the equation by backtracking how you find the roots.