Parabola Equation - Honors Algebra II?

Question

I was absent a day in my algebra II class...we covered parabolas. I think I have a pretty solid grasp on them, but one equation from my homework is giving me a ton of trouble.

I need to turn:
y[squared] - 8x = 0

Into this equation:
y = a(x-h)[squared] + k
OR
y = ax[squared]

From there, I can graph the parabola.

I've tried everything I can think of...completing the square, factoring, and nothing seems to work. Anything you can do to help, like explaining the steps, would be great. Thanks so much!

Answer 1

I would rearrange the formula to read: x=(y^2)/8. This is in the standard form: x-k=a(y-h)^2, where k and h are equal to 0. Therefore the vertex is at zero (since h=0 and k=0). Since it is an x= then it lies on its side and opens to the right. Since a=1/8 it opens very slowly. If a had been a large number it would have been steep and narrow when it opened.

Answer 2

Just rearange it
y^2 - 8x = 0
+8x to both sides of the equation
y^2 = 8x
square root each side
y= sqrt (8x)

Therefore, it IS a parabola. Since I took the square root, I get a plus OR minus. This gives you a parabola flipped over the line y=x

y= +-srt (8x)

email me if you have more issues

Answer 3

a million.) y = a(x-h)^2+ok y = 3(x-2)^2+5 the vertex in this equation is (h, ok). in case you adventure up the format on appropriate with the only on the backside, h=2 and ok=5 2.) y = x^2+8x-a million y = ax^2+bx+c in this equation, you utilize the equation -b/2a to discover the vertex. -8/2(a million) -8/2 = -4