Question

The vertex of the graph of a quadratic function is (3, 1) and the y-intercept is (0, 10). WHat is the equation of the function?

Answer 1

in vertex form it looks like
\[f(x)=a(x-3)^2+1\] the only thing you don't know is \(a\)
but since \((0,10)\) is on the graph you know
\[f(0)=a(0-3)^2+1=10\] you can solve that for \(a\)

Answer 2

how though i have no clue what im evan doin

Answer 3

how do you solve for \(a\) ?

Answer 4

yah

Answer 5

k lets do it step by step
what is \(0-3\) ?

Answer 6

-3

Answer 7

right
now what is \((-3)^2\) ?

Answer 8

9

Answer 9

k good
so now \[a(0-3)^2+1=10\] is
\[9a+1=10\] and we take two steps to solve for \(a\)
1) subtract 1
2) divide by 9

Answer 10

ok so 1?

Answer 11

yes
\(a=1\) so
\[f(x)=(x-3)^2+1\]

Answer 12

on account of you don't really need to write \[f(x)=1(x-3)^2+1\]

Answer 13

if you want, you could multiply out and combine like terms, or just leave it in "vertex form"

Answer 14

thats not one of the answers tho?

Answer 15

did you multiply out?
\[(x-3)^2+1=(x-3)(x-3)+1\] etc

Answer 16

yah

Answer 17

what did you get?

Answer 18

i dont remember...

Answer 19

ok thanks though satellite73 it was really helpfull