Question

Find a quadratic model for the set of values: (-2, -20), (0, -4), (4,-20) Show your work

Answer 1

@UsukiDoll

Answer 2

You need a system of equations

Answer 3

I don't feel well today. I woke up with a migraine (again)

Answer 4

Take some advil and drink water

Answer 5

$$ \Large{ y = ax^2 + bx + c
\\~\\
-20 = a(-2)^2 + b(-2) + c
\\-4 = a(0)^2 + b(0) + c
\\ -20 = a(4)^2 + b(4) + c

}$$

Answer 6

i honestly have no clue how to do this equation

Answer 7

Do you agree with what I have so far. I plugged in the values into a generic quadratic function.

Answer 8

I think so, i wouldn't know if you were wrong

Answer 9

Finding a quadratic model means fitting the points to a quadratic curve \( \large y = ax^2 + bx + c \) . So our job is to find the parameters \( \large a,b,c\)

Answer 10

@perl how do i find the a,b,c

Answer 11

We have these three equations so far. \[ \\~\\\Large {
-20 = a(-2)^2 + b(-2) + c
\\-4 = a(0)^2 + b(0) + c
\\ -20 = a(4)^2 + b(4) + c
}\]

Answer 12

the second equation gives us
-4 = 0 + 0 + c
-4 = c

Answer 13

So c=-4

Answer 14

Yes. Now back substitute.

Answer 15

We have these three equations so far. \[ \\~\\\Large {
-20 = a(-2)^2 + b(-2) + (-4)
\\-4 = a(0)^2 + b(0) + (-4)
\\ -20 = a(4)^2 + b(4) + (-4)
}\]

Answer 16

we don't need the second equation any longer

Answer 17

\[ \\~\\\Large {
-20 = a(-2)^2 + b(-2) + (-4)
\\ -20 = a(4)^2 + b(4) + (-4)
}\]

Answer 18

im writing as we go

Answer 19

\[ \\~\\\Large {
-20 = a(-2)^2 + b(-2) -4
\\ -20 = a(4)^2 + b(4) -4
\\~\\} \rm simplify \\~\\\Large {
-20 = 16a -2b -4
\\ -20 = 16a + 4b -4
\\~\\ \rm simplify ~ more\\~\\
-16 = 16a -2b
\\ -16 = 16a + 4b

}\]

Answer 20

Is that the final answer?

Answer 21

We can solve for `b`

Answer 22

So we haven't solved A or C yet?

Answer 23

we solved for c = -4

Answer 24

Ok so how do we solve for A and B than? Im sorry i really dont understand

Answer 25

@perl

Answer 26

we can solve that linear equation

Answer 27

how?

Answer 28

\[ \\~\\\Large {
-20 = a(-2)^2 + b(-2) -4
\\ -20 = a(4)^2 + b(4) -4
\\~\\} \rm simplify \\~\\\Large {
-20 = 16a -2b -4
\\ -20 = 16a + 4b -4
\\~\\ \rm simplify ~ more\\~\\
-16 = 16a -2b
\\ -16 = 16a + 4b
\\~\\ \rm add ~ the ~ two ~ equations\\~\\
\\-32 = 2b \\~\\
\\b = \frac{-32}{2}
}\]

Answer 29

So b=-16?

Answer 30

yes

Answer 31

So now we need to find A

Answer 32

y = -2x^2 + 4x -4 should be the answer.
I have an error somewhere above.

Answer 33

so b should really be b=4?

Answer 34

A=-2 B=4 C=-4?

Answer 35

yes

Answer 36

I have to go back and find the error now ;)

Answer 37

Awesome thank you so much!

Answer 38

but this method is sound if you want to try it again

Answer 39

if you ever*

Answer 40

yeah i prob will, i have a whole other algebra class ahead of me

Answer 41

\[ \\~\\\Large {
-20 = a(-2)^2 + b(-2) -4
\\ -20 = a(4)^2 + b(4) -4
\\~\\} \rm simplify \\~\\\Large {
-20 = \color{red}4a -2b -4
\\ -20 = 16a + 4b -4
\\~\\ \rm simplify ~ more\\~\\
-16 = \color{red}4a -2b
\\ -16 = 16a + 4b
\\~\\ \rm solve ~this~ system~ of~ equations\\~\\
\\ a = -2 ~ , b = 4
\\~\\ \rm backsubstitute
\\y = -2x^2 + 4x - 4
}\]

Answer 42

Another way to solve this problem is to note that your parabola has a y intercept of -4, because x=0 at the y intercept.
That reduces the problem of solving a linear system of 2 equations as we have above.