I have found out the coordinates are (-1,0)(4,0)(0,-8) but I can't find the turning point Thanks!
Last edited by agehayoshina; Dec 15th 2012 at 10:50 AM.
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the vertex, or 'turning point", of $\displaystyle f(x) = ax^2 + bx + c$ is at $\displaystyle x = \frac{-b}{2a}$
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!
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 $\displaystyle \color{red}y = f\left(\frac{3}{2}\right)$
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