# Parabolas and Other confusing equations

So here's the deal. I have three equations for three parabolas. I need to find the x and y inercepts, and a simplified radical to go with it.

Here are the equations.
Parabola one: y=(x-5)^2+6, Vertex: (5,6)
Parabola two: y=2(x-1)^2-4, Vertex: (1, -4)
Parabola three: y=-1(x+3)^2+8, Vertex: (-3, 8)

Does anyone know how to set this problem up and what equations I should use?

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Are you trying to find the x and y intercepts of each of the three parabolas? The x-intercept is when y = 0 and the y-intercept is when x = 0, so just plug away.

Are you trying to find the x and y intercepts of each of the three parabolas? The x-intercept is when y = 0 and the y-intercept is when x = 0, so just plug away.
Yes I'm trying to find the x and y intercepts of each of the three parabolas. But I don't really understand your method. Are you saying just to get each x an y alone and solve?

Yes.

Hmm, okay thanks

Being on the x & y intercepts refers to being on the x & y axis, respectively. If you want to be on the x-axis, what must the y-value be? Same goes for the y-int value.

that makes no sense, and it doesn't answer my question. I know what x and y intercepts are. I know that if I want to be on the x axis the y value has to be 0 and so forth.

I want to know an equation I can use to calculate them from those three equations.
Do I use the quadratic formula? or another quation?

Parabola one: y=(x-5)^2+6, Vertex: (5,6)

Show me the steps you would take to find the x-int for this problem.

There is no x intercept. It's above the x axis and it goes up.
but if i didn't, I would probably do the quadratic formula

In this case, in order to use the quadratic equation, you would need the values for a, b, and c. To get that, you would need to expand (x-5)2. That's absolutely possible. However, it is even easier in this case, because plugging in 0 for y would give you:

0 = (x-5)2 + 6 => -6 = (x-5)2

Do you see how you can solve for x very simply? In fact, if you had the equation in the form ax2 + bx + c and not in the vertex form in which they were given, one way of solving for the roots (x-intercepts), you may remember, is completing the square, which is what you did.

Moreover, the quadratic formula itself was figured out by completing the square for the general form ax2 + bx + c, just using all letters instead of numbers.

Ah I see.
Thanks. That was much more helpful than any of the other answers I got.

Ah I see.
Thanks. That was much more helpful than any of the other answers I got.
LOL ... wow, you need to read a little more into what others have suggested to you. Effort goes a long way, extends further than just solving a parabolic equation.

rocomath, I think you were extremely helpful. Thank you for your input.