Question

find the x-intercept of the parabola with vertex (-3,-14) and y intercept (0,13) in (x1,y1) (x2,y2) form using - (y-k)=a(x-h)^2

Answer 1

again?

Answer 2

Please see the attachment. This method works for whenever you have the vertex (h,k) and another point on the quadratic.

Answer 3

looks good to me

Answer 4

oh i see the instruction are not clear at all are they?

Answer 5

are you also tabbeyx33 because this is exactly what is posted there as well. i would just try to enter
\[y+14=3(x+3)^2\] because i think the instructions are left over from some other problem.

Answer 6

Oops, didn't finish the problem off.

Answer 7

no my name is priscila and i just need an asnswer to putt lol.

Answer 8

i have never heard of
\[(x_1,y_1), (x_2,y_2)\] form for a parabola so maybe if i watch jabberwock i will learn something

Answer 9

oh lord it says "find the x intercept" ok fine. start with
\[y=3(x+3)^2-14\], set y = 0 and solve!

Answer 10

\[0=3(x+3)^2-14\]
\[14=3(x+3)^2\]
\[\frac{14}{3}=(x+3)^2\]
\[x+3=\pm\sqrt{\frac{14}{3}}\]
\[x=-3\pm\sqrt{\frac{14}{3}}\]

Answer 11

a more traditional way to do this is write
\[y=3(x+3)^2-14\] and expand. you get
\[y=3(x^2+6x+9)-14\]
\[y=3x^2+18x+27-14\]
\[y=3x^2+18x + 13\]
set
\[0=3x^2+18x + 13\]
an use the quadratic formula to get
\[x=\frac{-9\pm\sqrt{42}}{3}\]

Answer 12

so x intercepts are
\[(\frac{-9-\sqrt{42}}{3},0)\] and
\[(\frac{-9+\sqrt{42}}{3},0)\]

Answer 13

that is the answer you need to put in and if you want a decimal use a calculator to get it

Answer 14

you sure?

Answer 15

i can try again

Answer 16

no is ok i put it in already & it was wrong :(
but is ok lol. ^_^

Answer 17

i got another one

Answer 18

yes i am sure. pretty sure anyway

Answer 19

fire away

Answer 20

nevermind i closed my test like a dumbass -____

Answer 21

on line class? dang

Answer 22

i need help with word problems.

Answer 23

ask

Answer 24

here you go.

Answer 25

can you see it?

Answer 26

no

Answer 27

did you attach it?