SOLUTION: Find the vertex and intercepts for the quadratic function and sketch the graph. y=x^2+2x-24 My thoughts: 0=x^2+2x-24....but then I'm not sure where I go from here. And I

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Question 425396: Find the vertex and intercepts for the quadratic function and sketch the graph.
y=x^2+2x-24

My thoughts:
0=x^2+2x-24....but then I'm not sure where I go from here. And I know once I find x, I need to find y?
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
the general vertex form of a quadratic is ___ y = a(x - h)^2 + k ___ (h,k) is the vertex and the "a" is related to the shape

separating out a perfect square ___ y = x^2 + 2x + 1 - 25 = (x + 1)^2 - 25

so the vertex is (-1,-25)

the x-intercepts (roots) are when y equals zero ___ factoring ___ 0 = (x + 6)(x - 4)

the y intercept is when x equals zero ___ y = (0)^2 + 2(0) - 24

this gives you four points to sketch the parabola
remember that the vertex is the "nose" of the parabola and it is on the axis of symmetry