Question

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

Answer 1

Since u need to show ur work....the general quadratic model is y=ax^2 + bx + c

Now we need to find the constants a, b, c ok?

Answer 2

How do we do that

Answer 3

U are given 3 points: (-2, -20), (0, -4), (4, -20)

So u can use them to solve for abc

Answer 4

I suck at this and I don't get it

Answer 5

I will do the first one for u, @Raylynngrace :)

y=ax^2 + bx + c for (-2, -20)
so
-20 = a*(-2)^2 + b*(-2) + c
-20 = 4a - 2b + c

ok?

Answer 6

Okay I get that now lol

Answer 7

Cool! Show me how u set up with the other two given points plz?

Answer 8

idk if I did this right lol but for the next one I got
y=ax^2+bx+c
-4=a*(0)^2+b*(0)
-4=a-b+c

Answer 9

good try but u missed something...it should be like

-4=a*(0)^2+b*(0) + c
-4=0+0+c
c=-4

hey we found c already!

Answer 10

Least I tried lol can you help me with the last one

Answer 11

nope u do it plz? i will help of cus :)

Answer 12

y=ax^2+bx+c
-20=a*(4)^2+b*(4)
-20=16a+4b

Answer 13

Did I get this one wrong to @sdfgsdfgs

Answer 14

@raylynngrace hahahaa u keep forgetting the c term....so it should be:

-20=a*(4)^2+b*(4)+c
-20=16a+4b+c

now u got c=-4 already and from the first point:
-20 = 4a - 2b + c

U can solve for a and b :)

Answer 15

@Raylynngrace
This will help you a bit. Try !!

http://www.softschools.com/math/algebra/quadratic_functions/quadratic_function_with_three_points/

Answer 16

Thank you both for your help @sdfgsdfgs @JoshKoikkara