Question

Find the x-intercepts of the parabola with vertex (-4,2) and y-intercept (0,-30). Write your answer in this form: (x1, y1),(x2, y2). If necessary, round to the nearest hundredth

Answer 1

First find the equation of this parabola, you have the vertex and the y-intercept:

\(y=a(x-h)^2+k\\
-30=a(0-(-4))^2+2\\
-30=a(4)^2+2\\
16a+2=-30\\
-30-2=16a\\
-32=16a\\
a=-32/16\\
a=-2\)

So the equation of this parabola would be \(y=-2(x+4)^2+2\)

Now to find the x-intercepts, replace y by 0 and solve for x.

Answer 2

I got x=2...

Answer 3

There's two of them. Remember, a square root always gives 2 answers, e.g.: \(\sqrt{25}=-5 ~and~ 5\)

Answer 4

I put it in as (-5,0), (5,0)
It was wrong..

Answer 5

It's an example D:

Answer 6

oh...
Well the one I'm on now has a vertex of (-7,45) nd y-intercept of (0,-200)

Answer 7

\( 0=−2(x+4)^2+2\\
-2=-2(x+4)^2\\
1=(x+4)^2\\
\pm \sqrt{1}=x+4\)

So we have -1 = x+4; x=-1-4=-5

and 1=x+4; x=1-4=-3

Roots at (-5,0) and (-3,0)

Answer 8

That was wrong too.. I think I might be typing it wrong...

Answer 9

Vertex of (2, 13) and y-intercept of (0, 5)

Answer 10

That one, is the previous question...

Answer 11

...I might throw my computer.. D: I'm dumb
lgogbe

Answer 12

I got it now, thanks!

Answer 13

You are welcome :)

Answer 14

Can you help me with another one? It's my last one an I need to get it right.

Find the x-intercepts of the parabola with vertex (4,-1) and y-intercept (0,15). Answer in this form: (x1,y1),(x2,y2)