Question 232715: can you please help be with these problems.find the (x,y)value of the vertex of the parabola, y=x^2-16
find the (x,y0value of the vertex of the parabola, y=x^2+6x+5
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! The "vertex form" is:
y= a(x-h)^2+k
where (h,k) is the vertex
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find the (x,y)value of the vertex of the parabola,
y=x^2-16
Rewriting to vertex form we have:
y=(x-0)^2+(-16)
Therefore, the vertex is at (0, -16)
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find the (x,y) value of the vertex of the parabola,
y=x^2+6x+5
Completing the square:
y=(x^2+6x+__)+5
y=(x^2+6x+9)+5-9
y=(x+3)^2 - 9
Rewriting to vertex form we have:
y=(x-(-3))^2 + (-9)
Therefore, the vertex is at (-3, -9)
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