You can put this solution on YOUR website! 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.
You can put this solution on YOUR website! If we complete the square on this, we get
y = (x+1)^2 -4
So, the vertex is (-1.4)
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The x-intercepts occur when y=0, so we get
0 = (x+1)^2 - 4
adding 4 we get
4 = (x+1)^2
taking a square root, we get
+-2 = x+1
subtracting 1, we get two answers
X = 1
and
x = -3
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The y-intercept occurs when x=0, so we get
y = (0+1)^2 - 4
y = 1^2 - 4
y = -3
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the graph is
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