Question

find the x-intercepts of the parabola with vertex (-3,16) and y-intercept (0,-20). Write your answer in this form: (x1,y1),(x2,y2), round to the nearest hundredth.

Answer 1

Hint: y = a(x - h)^2 + k, where (h, k) is the vertex. Then, substitute to find a, and solve the equation.

Answer 2

I dont know how.

Answer 3

@imilee Firstly substitute (h, k) with the vertex value you have.

Answer 4

y=a(x-3)^2+16

Answer 5

Now, substitute the value of the y intercept into the equation.

Answer 6

y=0(20-3)^2+16?

Answer 7

@imilee a stays intact, because that's the variable you're looking for.

0 = a(20 - 3) ^2 + 16. Then solve for a.

Answer 8

How?
a=1^2

Answer 9

@imilee Ah, I see what you did wrong: When you substitute, notice the signs: x - h means x - (-3), not x - 3.
Your equation should be y=a(x+3)^2+16 - notice the + in (x+3).

So, substitute values and you get 0=a(20+3)^2+16

Answer 10

soo a=39^2?

Answer 11

@imilee Nope. Did you try simplifying and solving the equation?

Answer 12

0=a(20+3)^2+16
a=23^2+16
a=39^2
thats what I did.

Answer 13

No. a(20+3)^2 is one term, they must stick together. After simplifying you should get (a)(23^2) = 529a.

Answer 14

but how do I put it as this form: (x1,y1),(x2,y2)?

Answer 15

You must first solve for a, then solve for the equation to find the x intercepts.

Answer 16

how do I solve (a)(23^2) = 529a.

Answer 17

You are solving for 0=a(20+3)^2+16. I was saying that (a)(23^2) after simplifying equals to 529a.

Answer 18

then what do I do with 529a

Answer 19

@imilee - Simplify and solve the equation.