# Finding the turning point for the curve f(x)2x^2-6x-8

• Dec 15th 2012, 10:46 AM
agehayoshina
Finding 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 AM
skeeter
Re: 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 AM
agehayoshina
Re: 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 AM
skeeter
Re: Need help finding the turning point for the curve f(x)2x^2-6x-8
Quote:

Originally Posted by agehayoshina
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?

you have miscalculated the x-value of the vertex.

$\displaystyle f(x) = 2x^2 - 6x - 8$

$\displaystyle x = \frac{-b}{2a} = \frac{-(-6)}{2(2)} = \frac{3}{2}$

the y-value of the vertex is $\displaystyle \color{red}y = f\left(\frac{3}{2}\right)$