Question

Real quick, last one! Find the x-intercepts of the parabola with the vertex (-1,2) and y-intercept (0,-3)

Answer 1

\[f(x)=a(x-h)^2+k\]
where (h,k) is vertex
but we vertex is (-1,2) => h=-1 and k=2
so we have
\[f(x)=a(x-(-1))^2+2\]
now we are given f(0)=-3 use this to find a

Answer 2

???

Answer 3

Solve

\[-3=a(0+1)^2+2\]for a

Answer 4

Ok hang on

Answer 5

that will give you the quadratic... then you can find the roots.

Answer 6

Sorry I am completely awful at looking at something and turning it into the problem I need

Answer 7

a=-5

Answer 8

yes

Answer 9

Where do I plug that in?

Answer 10

into

\[f(x)=a(x+1)^2+2\]

Answer 11

Ok hang on

Answer 12

then solve \[f(x)=0\]

for x

Answer 13

What do I do to f(x)=-5(x+1)^2+2 find the vertex or what?

Answer 14

you want the x-intercepts

so solve f(x)=0 for x

Answer 15

solve

\[-5(x+1)^2+2=0\]

Answer 16

0.37 and -1.63 is what I got for the x intercepts...

Answer 17

is that correct?

Answer 18

-.37

Answer 19

\[-1\pm\frac{\sqrt{10}}{5}\]

Answer 20

(-.367544467966,0)
(-1.63245553203,0)

Answer 21

Thank you so much for helping me

Answer 22

np