I have found out the coordinates are (-1,0)(4,0)(0,-8) but I can't find the turning point :(

Thanks!

Printable View

- Dec 15th 2012, 10:46 AMagehayoshinaFinding the turning point for the curve f(x)2x^2-6x-8
I have found out the coordinates are (-1,0)(4,0)(0,-8) but I can't find the turning point :(

Thanks! - Dec 15th 2012, 10:50 AMskeeterRe: Need help finding the turning point for the curve f(x)2x^2-6x-8
the vertex, or 'turning point", of $\displaystyle f(x) = ax^2 + bx + c$ is at $\displaystyle x = \frac{-b}{2a}$

- Dec 15th 2012, 10:54 AMagehayoshinaRe: Need help finding the turning point for the curve f(x)2x^2-6x-8
Hmm this is till kind of confusing. If I plug the numbers in i get -6/2(2) =-1.5. That's x. How do I find y?

Thanks! - Dec 15th 2012, 11:53 AMskeeterRe: Need help finding the turning point for the curve f(x)2x^2-6x-8