SOLUTION: How do you put [x=y^2+8y+13] into vertex form and how do you graph it?

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Question 82783: How do you put [x=y^2+8y+13] into vertex form and how do you graph it?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How do you put [x=y^2+8y+13] into vertex form and how do you graph it?
Square the y-terms to get:
x=y^2+8y+16+13-16
x=y^2+8y+16-3
x+3 = (y+4)^2
Vertex at (-3,-4)
x-intercept at 13
y-intercepts:
y = [-8+-sqrt(8^2-4*13)]/2
y = [-8+-2sqrt3]/2
y = (-4+sqrt3) or y = -4-sqrt3
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The parabola opens to the right; has vertes, x-intercept, and y-intercepts
as indicated.
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Cheers,
Stan H.