Question

27

Answer 1

i'm sure there's some equation that if you can transform your quadratic equation into, but i would just expand what you have, take the derivative, and find where the derivative is 0. that's your vertex

Answer 2

What are the values of x that make y = 0? The vertex will be located at the value of x which is equidistant to those two values of x, and you can use the formula to find its value.

Answer 3

If you expand this into the form \(y = ax^2+bx+c\), you can find the x value of the vertex by evaluating \(-\frac{b}{2a}\) as well.

Answer 4

The first approach works because the parabola is symmetric about a line going through the vertex. The second approach is the canned version of the calculus approach suggested by @robz8

Answer 5

yeah my way wouldn't be helpful if you are not in calculus yet!

Answer 6

It's of course easy to find the values of x where the function equals 0:

\[-(x-3)(x+1) = 0\]if \((x-3) =0\) or if \((x+1) = 0\)

Answer 7

i found an equation y=a(x-r1)(x-r2) x=r1+r2/2

Answer 8

that's what my first post said.