SOLUTION: find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2 - 8x +3

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2 - 8x +3      Log On


   

Question 399686: find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2 - 8x +3
Found 2 solutions by Alan3354, nerdybill:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
y=-2x^2 - 8x +3
The eqn of the axis of symmetry is x = -b/2a
x = 8/-4
x = -2
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The vertex is (x,f(x))
f(-2) = -2*4 + 16 + 3 = 11
= (-2,11)

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
y=-2x^2 - 8x +3
.
Axis of symmetry:
x = -b/(2a)
x = -(-8)/(2(-2))
x = 8/(-4)
x = -2
.
plug above into:
y=-2x^2 - 8x +3
y=-2(-2)^2 - 8(-2) +3
y=-2(4) - (-16) +3
y=-8 + 16 +3
y = 11
.
vertex is at (-2, 11)