<PRE><B>Question 561628</B>
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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