SOLUTION: Please show me how to solve the quadratic equation, I need to get the vertex, axis of sym., and y-intercept to graph. 2y=X^2+4X+5

Question 138938: Please show me how to solve the quadratic equation, I need to get the vertex, axis of sym., and y-intercept to graph.
2y=X^2+4X+5

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I need to get the vertex, axis of sym., and y-intercept to graph.
2y=X^2+4X+5
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Solve for y:
y = (1/2)x^2 + 2x + (5/2)
You have a quadratic with a= 1/2 ; b= 2, c= (5/2)
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Vertex occurs at x = -b/2a = -2/(2(1/2)) = -2
f(-2) = (1/2)(1/4) - 4 + 5/2 = -11/8
Vertex at (-2,-11/8)
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Axis of symmetry is a vertical line thru the vertex: x = -2
--------------------
y-intercept at f(0) = (5/2)
--------------------

Cheers,
Stan H.
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