Question

PLEASE HELP!!!

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

Answer 1

\(\large\color{#000000}{\displaystyle y= ax^2+bx+c }\)

is what the quadratic equation looks like.

Answer 2

I got disconnected, apologize.

Answer 3

ok and then what? i have no idea on how to do this :(

Answer 4

You are given 3 points so plug them in (individually) to solve for a, b, & c.

Answer 5

oh no worries its fine

Answer 6

plug in the points;

For (-2,-20),
\(\large\color{#000000}{\displaystyle -20= a(-2)^2+b(-2)+c }\)

For (0,-4),
\(\large\color{#000000}{\displaystyle -4= a(0)^2+b(0)+c }\)

For (4,-20),
\(\large\color{#000000}{\displaystyle -20= a(4)^2+b(4)+c }\)

Answer 7

Simplify and solve the system...

Answer 8

(note: it is very easy to obtain the c from the 2nd equation)

Answer 9

i dont understand

Answer 10

each of those points
(-2, -20), (0, -4), (4, -20)
is on the parabola,

therefore, you can plug them into the parabola to solve for a, b and c.
I plugged everything for you you just need to simplify and solve.

Answer 11

let's go with the first equation:
\(\large\color{#000000}{\displaystyle -20= a(-2)^2+b(-2)+c }\)

can you simplify this one?

Answer 12

ok i get that but how would i simplify them?

Answer 13

(-2)² = ?

Answer 14

4?

Answer 15

yes

Answer 16

\(\large\color{#000000}{\displaystyle -20= a(-2)^2+b(-2)+c }\)

\(\large\color{#000000}{\displaystyle -20= 4a-2b+c }\)

Answer 17

\(\large\color{#000000}{\displaystyle -20= 4a-2b+c }\) is your first equation.. ok?

Answer 18

ok

Answer 19

Now simplify the second equation

Answer 20

\[-4=a+b+c\]
?

Answer 21

@SolomonZelman

Answer 22

\(\large\color{#000000}{\displaystyle -4= a(0)^2+b(0)+c }\)

what is 0²?

Answer 23

0
right?

Answer 24

yes, and what is a times 0?

Answer 25

0

Answer 26

what is b•0 ?

Answer 27

0?

Answer 28

yes

Answer 29

So what does the second equatrion simplify to?

Answer 30

-4=a+c?

Answer 31

almost right

Answer 32

we said that 0²=0, and a•0=0, so what happens to the "a(0)²" component?

Answer 33

it goes away?

Answer 34

yes, so what is your second equation?

Answer 35

umm...
-4=c?

Answer 36

yes

Answer 37

\(\large\color{#000000}{\displaystyle -20= 4a-2b+c }\)
\(\large\color{#000000}{\displaystyle -4= c }\)

now lets do the 3rd equation.

Answer 38

\(\large\color{#000000}{\displaystyle -20= a(4)^2+b(4)+c }\)

Answer 39

4² = ?

Answer 40

8?

Answer 41

wait no 16

Answer 42

yes
4² = 4•4 = 16

Answer 43

so your third equation, when simplified, would be?

Answer 44

\[-20=16a+4b+c\] ?

Answer 45

fabulous!

Answer 46

^-^

Answer 47

So we have,

\(\large\color{#000000}{\displaystyle -20= 4a-2b+c }\)
\(\large\color{#000000}{\displaystyle -4= c }\)
\(\large\color{#000000}{\displaystyle -20=16a+4b+c }\)

Answer 48

Note (again), that in 2nd equation you are given the value of c explicitely:
c=-4

And you can use that to solve the 1st and 2rd equations for "a" and "b".

Answer 49

plug in c=-4, into equation 1 and into equation 3.

Answer 50

so...
\[-20=4a-2b+-4\]
?

Answer 51

yes, the first equation...
and that would be better written as,
\(\large\color{#000000}{\displaystyle -20=4a-2b-4 }\)

Answer 52

Now, plug in c=-4, into the 3rd equation, please..

Answer 53

\[-20=16a+4b+-4\]
?

Answer 54

YES!
and that would be better written as?

Answer 55

@SolomonZelman ?