SOLUTION: I needed to find the vertex of this parabola: y = -x^2 - 3x - 2 so what I did was: 0 = -x^2 - 3x - 2 x = -1, x = -2 Two x-intercepts: (-1, 0) and (-2, 0) y = -0^2 -

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: I needed to find the vertex of this parabola: y = -x^2 - 3x - 2 so what I did was: 0 = -x^2 - 3x - 2 x = -1, x = -2 Two x-intercepts: (-1, 0) and (-2, 0) y = -0^2 -       Log On

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Question 58609: I needed to find the vertex of this parabola: y = -x^2 - 3x - 2
so what I did was:
0 = -x^2 - 3x - 2
x = -1, x = -2
Two x-intercepts: (-1, 0) and (-2, 0)
y = -0^2 - 3(0) - 2 = -2
One y-intercept: (0, -2)
x = 1/2(-1 - 2) = -3/2
y = -3/2
y = -3/2 - 2)-3/2) - 2 = 1.5
Vertex is -3/2, 1.5
Is this correct?

Answer by funmath(2873) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex of this parabola: y = -x^2 - 3x - 2
You got the x value right, but you're doing it in such a way that if you cannot factor it, you won't find it. Therefore, I'm going to show you the sure fire way of finding the x value of the vertex, no matter what and then we'll see what went wrong for the y-value:
The formula for finding the x-value of the vertex of a parabola is highlight%28x=-b%2F2a%29, for a quadratic equation written in standard form:highlight%28y=ax%5E2%2Bbx%2Bc%29
Your a=-1, b=-3, and c=-2
x=-%28-3%29%2F%282%28-1%29%29
x=3%2F-2
x=-3%2F2=-1.5
Substitute that value into the equation for x and solve for y:
y=-%28-1.5%29%5E2-3%28-1.5%29-2
y=-2.25%2B4.5-2
y=.25
The vertex is: (x,y)=(-1.5,.25)
Happy Calculating!!!