Graphing a parabola for a calculus problem

[hotrocks007]

I am graphing a parabola for a calculus problem. I understand how you factor and use the two x values as your x-intercepts, but I'm not sure how this one would be graphed.

y=4x^2 - 25. I understand how the vertex of the parabola would be at negative 25, but I have no clue what to do woith the 4x^2.
Help would be appreciated. thanks.

 

[LeonhardEuler]

Well, like you said factoring and getting the two x-intercepts is a start. You also want to know the turning point. If you don't remember the formula for that, just remember that the derivative is zero at the turning point and solve.

 

[mathmike]

factoring you get (2x-5) (2x+5) set these equal to zero

 

[HallsofIvy]

By the way, the vertex is not "at negative 25"- that's just the y-value of the point. The vertex is at (0, -25).