Question

help @jim_thompson5910 @jdoe0001

Answer 1

with? mowing the lawn? going to the supermarket for milk? getting a subprime rate loan from a fishy bank?

Answer 2

in general

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

has the vertex (h,k)

so you plug in the vertex to get

x = a(y-(-1))^2 + (-3)

then you plug in the point (4,0) and then solve for 'a' to get your answer

Answer 3

how do you plug it in?

Answer 4

x = 4, y =0

x = a(y-(-1))^2 + (-3)

4 = a(0-(-1))^2 + (-3)

now solve for 'a'

Answer 5

x=4?

Answer 6

yes

Answer 7

x=4 is a GIVEN, as you read the sentence :)

Answer 8

and y=0 is also a GIVEN value

Answer 9

so, it's like christmas time, you're getting lots of GIVEN ones hehe

Answer 10

can you help me figure it out?

Answer 11

4 = a(0-(-1))^2 + (-3)

4 = a(0+1)^2 + (-3)

4 = 1a - 3

4 = a - 3

....
....

a = ???

Answer 12

3?

Answer 13

no please don't just make random guesses

Answer 14

4 = a - 3

a = ??

Answer 15

im confused :(

Answer 16

do i add the 4 and the 3?

Answer 17

let's see, for a parabola whose vertex is (-3, -1)
$$
(\color{red}{h},\color{blue}{k}) \implies (\color{red}{x},\color{blue}{y}) \implies (\color{red}{-3},\color{blue}{-1})\\
\text{a point value given is } (4,0)\\
4=a(0-\color{red}{(-3)})^2+\color{blue}{(-1)}\\
$$
solve for "a"
what would that give you?

Answer 18

yes you need to add 3 to both sides

Answer 19

that gave me 7 though

Answer 20

so a = 7

Answer 21

that's the coefficient for y^2

Answer 22

idk?

Answer 23

where are you stuck? you have the answer

Answer 24

oo so it is 7?

Answer 25

yes