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

A quadratic function:
[tex]y=ax^2+bx+c[/tex]

First, take the point (0,-4) and plug the values (x,y) into the equation:
[tex]-4=a \times 0^2+b \times 0 +c \\ -4=c[/tex]

So the equation is [tex]y=ax^2+bx-4[/tex].

Now plug the values of the other two points into the equation and set up a system of equation:
[tex]-20=a \times (-2)^2+b \times (-2)-4 \\ -20=a \times 4^2+b \times 4-4 \\ \\ -20+4=4a-2b \\ -20+4=16a+4b \\ \\ -16=4a-2b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \div 2 \\ -16=16a+4b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\ \\ -8=2a-b \\ \underline{-4=4a+b} \\ -12=6a \\ \frac{-12}{6}=a \\ a=-2 \\ \\ -8=2a-b \\ -8=2 \times (-2)-b \\ -8=-4-b \\ -8+4=-b \\ -4=-b \\ b=4[/tex]

The function is:
[tex]\boxed{y=-2x^2+4x-4}[/tex]

carlosego carlosego · Answer 2 · 2018-07-26T00:50:17+02:00

For this case, the quadratic function in its generic form is given by:

[tex] y = ax ^ 2 + bx + c
[/tex]

We must find the values of the coefficients.

For this, we evaluate the given points.

For (0, -4):

[tex] -4 = a (0) ^ 2 + b (0) + c

c = -4
[/tex]

For (-2, -20):

[tex] -20 = a (-2) ^ 2 + b (-2) + c

4a - 2b - 4 = - 20

4a - 2b = - 20 + 4

4a - 2b = - 16
[/tex]

For (4, -20):

[tex] -20 = a (4) ^ 2 + b (4) + c

16a + 4b - 4 = - 20

16a + 4b = - 20 + 4

16a + 4b = - 16
[/tex]

Therefore, for the values of a and b we have the following system of equations:

[tex] 4a - 2b = - 16

16a + 4b = - 16
[/tex]

Resolving graphically (see attached image) we have:

[tex] a = - 2

b = 4
[/tex]

Then, the quadratic model is:

[tex] y = -2x ^ 2 + 4x - 4
[/tex]

Answer:

a quadratic model for the set of values is:

[tex] y = -2x ^ 2 + 4x - 4 [/tex]