SOLUTION: Given y=x^2+3x-28. How do I algebraically find: The vertex, The x and y intercepts, and write the equation in vertex and factor form. This makes absolutely no since to me at all, a

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Given y=x^2+3x-28. How do I algebraically find: The vertex, The x and y intercepts, and write the equation in vertex and factor form. This makes absolutely no since to me at all, a      Log On


   



Question 561628: Given y=x^2+3x-28. How do I algebraically find: The vertex, The x and y intercepts, and write the equation in vertex and factor form. This makes absolutely no since to me at all, and I'm super confused. PLEASE HELP ME!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Given y=x^2+3x-28. How do I algebraically find: The vertex, The x and y intercepts, and write the equation in vertex and factor form.
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Complete the square:
y = x^2+3x + 9 -9-28
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y = (x+3)^2-37
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Vertex: (-3,-37)
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y-intercept:
Use the form y = x^2+3x-28
Let x = 0, then y = -28
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x-intercepts:
Solve x^2-3x-28 = 0
Factor:
(x-7)(x+4) =
x = 7 ; x = -4
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Cheers,
Stan H.
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