Question 259544
Graph y = x^2+2x–3 and give the vertex and 
x and y intercepts.
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Put the equation in vertex form.
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x^2 + 2x +? = y+3+?
x^2 + 2x +1 = y + 4
(x+1)^2 = y +4
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Vertex: (-1,-4)
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y = x^2+2x–3
x-intercepts: 
Let y = 0 and solve for "x"
x^2 + 2x -3 = 0
(x+3)(x-1) = 0
x = -3 or x = 1
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y-intercept:
Let x = 0 and solve for "y":
y = 0^2 + 2*0-3
y = -3
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Cheers,
Stan H.