R rn50nurse New member Joined Aug 5, 2007 Messages 22 Aug 8, 2007 #1 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!

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!

G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Aug 8, 2007 #2 Finding the vertex is easy. You have \(\displaystyle \L\\x^{2}-2x-8\) a=1, b=-2, c=-8 To find the vertex, \(\displaystyle \L\\x=\frac{-b}{2a}\) \(\displaystyle \L\\y=c-\frac{b^{2}}{4a}\)

Finding the vertex is easy. You have \(\displaystyle \L\\x^{2}-2x-8\) a=1, b=-2, c=-8 To find the vertex, \(\displaystyle \L\\x=\frac{-b}{2a}\) \(\displaystyle \L\\y=c-\frac{b^{2}}{4a}\)

R rn50nurse New member Joined Aug 5, 2007 Messages 22 Aug 10, 2007 #3 I think I had a "brain fart" thanks for the input-- :lol: