Question

I AM AWARDING MEDALS!


Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of
(– 2, – 16), and x intercepts at x = -6 and x = 2. (Do not include the negative sign in your answer.)

Answer 1

as the kids say "tmi"

Answer 2

Hey its @julian25 , you gonna help

Answer 3

Oh, sorry @sa

Answer 4

@satellite73 *

Answer 5

yes give a minute

Answer 6

you can't?

Answer 7

lol

Answer 8

how do you think i got this nice green bike?

Answer 9

auction

Answer 10

lololol hahaha!

Answer 11

there are many different ways you can get a solution here
one way is to start by writing
\[y=a(x+6)(x-2)\]

Answer 12

Why do you do that

Answer 13

then since you have \((-2,-16)\) on the graph you can solve for \(a\) by writing
\[-16=(-2+6)(-2-2)\] and solve

Answer 14

if you know the zeros, you know how it factors

Answer 15

the other way it to write it in "vertex" form as
\[y=a(x+2)^2-16\]

Answer 16

y=a(x-h)^2+k
(h,k) is the vertex and for x and y we can use one of the x intercepts. Let's choose (2,0)
0=a(2+2)^2-16
0=16a-16
a=1

y=1(x+2)^2-16 is the required equation. If you want it in a different form, it can be rearranged

y=x^2+4x+4-16
y=x^2+4x-12

Answer 17

again you can solve for \(a\) by putting
\[0=a(2+2)^2-16\]

Answer 18

ahh that is what @FriedRice did above

Answer 19

as you can see you have too much information
one method required only the vertex and one point
the first method i wrote used both zeros, but ignored the fact that \((-2,-16)\) was the vertex

Answer 20

But wait I don't g et what to do? @satellite73

Answer 21

you are looking for a quadratic with three features
1) the vertex is \((-2,-16)\)
2) one of the zeros is \(x=2\) i.e. \((2,0)\) is on the graph
3) another zero is \(x=-6\) so \((-6,0)\) is on the graph

Answer 22

so far so good?

Answer 23

yes

Answer 24

we can do it as @FriedRice did above
write the quadratic in vertex form as
\[y=a(x+2)^2-16\] and the only number you do not know is \(a\)

Answer 25

but you know that if \(x=2\) you must have \(y=0\) because you are told that one of the zeros is \(2\)

Answer 26

So 0=a(2+2)^2-16

Answer 27

so replace \(y\) by \(0\) and \(x\) by \(2\) and get the equation
\[0=a(2+2)^2-16\] solve for \(a\) via
\[0=16a-16\\
16=16a\\
1=a\]

Answer 28

yes, what you said

Answer 29

So 0=0 if you work through

Answer 30

now we know that \(a=1\) so it is
\[y=(x+2)^2-16\] which you can expand if you like

Answer 31

no, not \(0=0\) but rather \(a=1\)

Answer 32

yes, watch my work
0=(2+2)^2-16
parentheses
0=4^2-16
exponents
0=16-16
subtraction
0=0

Answer 33

@satellite73