SOLUTION: I need help in finding the coordinates of the vertex for the parabola defined by the given quadratic function f(x) = -x^2 -4x + 8

Question 489785: I need help in finding the coordinates of the vertex for the parabola defined by the given quadratic function f(x) = -x^2 -4x + 8
Found 2 solutions by lwsshak3, MathTherapy:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I need help in finding the coordinates of the vertex for the parabola defined by the given quadratic function f(x) = -x^2 -4x + 8
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Given function is a parabola of the standard form: -A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. A is a multiplier which affects the steepness or slope of the parabola. The minus sign of the lead coefficient means that the parabola opens downwards, that is, it has a maximum.
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Finding the coordinates of the vertex:
f(x) = -x^2 -4x + 8
completing the square
y=-(x^2+4x+4)+8+4
y=-(x+2)^2+12
Ans:
Coordinates of the vertex: (-2,12)
See graph below as a visual check on the answer:
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Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
I need help in finding the coordinates of the vertex for the parabola defined by the given quadratic function f(x) = -x^2 -4x + 8

To find the axis of symmetry, or the x-coordinate of the vertex, we use the formula, , where a = - 1, and b = - 4

That results in: x = = = - 2

Now, substituting - 2 for x in the equation, we get: f(x) or y = ------ f(x) or y = - 4 + 8 + 8 ----- y = 12

Since the x and y-coordinates are - 2, and 12, respectively, then the coordinates of the vertex of the parabola are (, )

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