vertex of a parabola

[Guest]

can y'all help me? I just started school and I didnt do too well in pre-calculus. So can someone tell me how to find the vertex of a parabola.

eg y=x^2 -8x+13 HELP ME!!!!

 

[tkhunny]

I just started school and I didnt do too well in pre-calculus. So can someone tell me how to find the vertex of a parabola.
y=x^2 -8x+13

"FORM" is the answer. Typically, this is one variety:

(y-k) = 4p*(x-h)²

The Vertex is at (h,k). Make your equation look like that and you're done. Completing the Square is a good way to do it.

y=x^2 -8x+13
y - 13 = x^2 - 8x

8/2 = 4
4^2 = 16

y - 13 + 16 = x^2 - 8x + 16
(y+3) = (x-4)^2

Vertex: (4,-3)

If you get to use Calculus, the x-value can be tracked down much more quickly:

y=x^2 -8x+13

First Derivative:

(dy/dx = 2*x - 8

2*x - 8 = 0
x = 4 <== There it is.

Note: It's a bit trickier with parabolas having axes parallel to the x-axis.

 

[soroban]

Hello, badwithcalchica12!

There is a <u>formula</u> for this problem . . .

The parabola, .y = ax² + bx +c, .has its vertex at: .x = -b/2a

Find the vertex of: y = x² - 8x + 13
.

Your parabola has: .a = 1, b = -8, c = 13.

. . Then: .x .= .-(-8)/2(1) .= .4

. . . And: .y .= .4² - 8·4 + 13 .= .-3

Therefore, the vertex is: .(4,-3)