SOLUTION: i need help finding the axis of symmetry and vertex of the function f(x)=-2x^2, im confused because it doesnt give me B nor C only A.

Question 420127: i need help finding the axis of symmetry and vertex of the function f(x)=-2x^2, im confused because it doesnt give me B nor C only A.
Found 2 solutions by dnanos, Earlsdon:
Answer by dnanos(83) (Show Source): You can put this solution on YOUR website!

The symmetry axis is y'y and the vertex is the origin O(0,0).
This result can be proved also because of the formula of this function f(x)=, where a=-2...
(i)This x squared gives f(x)=f(-x) ,that is for every x the symmetric points
(x,f(x)) and (-x,f(-x)) belong to the graph.[these points are (x,f(x),(-x,f(x))and we know the are y'y-symmetric as any two points (a,b) and (-a,b) are.]
(ii)a=-2 that is a<0
so ,that is the maximum value of f(x)=0 occurs when x=0,and (x,y)=(0,0) is the vertex of the parabola.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
When you compare the given equation: with the general form for a quadratic equation: , you can see that: a = -2, b = 0, and c = 0.
The x-coordinate of the vertex is given by:
and, since b= 0, the x-coordinate is x = 0.
Substitute x = 0 into the given equation: to get:

The vertex is at (0, 0)
The graph of the given equation is a parabola that opens downward (negative coefficient of the term.
The equation of the axis of symmetry is or which is the y-axis.
Here's the graph of the given equation:

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