Quadratic eqns: y=X^2-2x-8 and y=-2x^2-x+2

[rn50nurse]

y=X^2-2x-8 and y=-2x^2-x+2

I am lost how to start this to find the vertex and points for the parabola. anyone Help? thanks!

 

[galactus]

Finding the vertex is easy.

You have \(\displaystyle \L\\x^{2}-2x-8\)

a=1, b=-2, c=-8

To find the vertex, \(\displaystyle \L\\x=\frac{-b}{2a}\)

\(\displaystyle \L\\y=c-\frac{b^{2}}{4a}\)

 

[rn50nurse]

I think I had a "brain fart" thanks for the input-- :lol: