Question

write the equation for each parabola with the given conditions:
Vertex(-3,-4); the graph passes through the point (1,4).
X Intercept of -2 and 1. then find the vertex

Answer 1

I need all the steps pleased!!!!!!!!!!!

Answer 2

Is the second line one problem?

Answer 3

no those are the given conditions you have to find an equation for each parabola
write the equation for each parabola with the given conditions:
A. Vertex(-3,-4); the graph passes through the point (1,4).
B. X Intercept of -2 and 1. then find the vertex

Answer 4

Oh so they're two different ones.

ok well for a, you know the vertex form of a parabola is

y=a(x-h)^2+k

where (h,k) is the vertex, and since you have another point on the graph, you can plug in the vertex and x and y to solve for a

Answer 5

SO I KNOW I PLUG IN A(X+3)^2 -4 THEN AFTER THAT WHAT DO I DO?

Answer 6

Well you have to plug in the other point so you can solve for a..

4=a(1+3)^2-4

Answer 7

THEN DO I DISTRIBUTE IT SO IT'LL GIVE ME 4=A(25-4?

Answer 8

Do the parentheses first

4=16a-4
8=16a
1/2=a

So, now that you know that you can say the equation is

y=1/2(x+3)^2-4

Answer 9

THIS IS WHAT I DID
4=A(1+4)^2-4
4=A(5)^2-4
4=A(25)-4
THEN IDK WHAT TO DO

Answer 10

Well its a 3 inside the parentheses, not a 4

Answer 11

HOW DID YOU GET 16?

Answer 12

1+3=4
4^2=16

Answer 13

ill do part 2 of the problem then

Answer 14

BUT ITS A -3 NOT A +3?

Answer 15

The sign changes because its (x-h)

Answer 16

THEN YOU ONLY DIVIDE 8 BY 16 = 1/2 RIGHT?

Answer 17

Assuming that this is a normal parabola without any coefficients in front of the x values...... you know your roots, since they are they x-intercepts... with that... you get...

\[(x+2)(x-1) = x^{2} + x-2\]
which you can find the x-value of the vertex.....

Answer 18

or you can get your x-value of the vertex since the x-value of the vertex has to be exactly between your roots.... since a parabola is a symmetrical function over it's center x-value or the x-value of the vertex