itze1213onofre
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# Find a quadratic model for the set of values: (-2, -20), (0, -4), (4,-20) Show your work

A quadratic function: $y=ax^2+bx+c$ First, take the point (0,-4) and plug the values (x,y) into the equation: $-4=a \times 0^2+b \times 0 +c \\ -4=c$ So the equation is $y=ax^2+bx-4$. Now plug the values of the other two points into the equation and set up a system of equation: $-20=a \times (-2)^2+b \times (-2)-4 \\ -20=a \times 4^2+b \times 4-4 \\ \\ -20+4=4a-2b \\ -20+4=16a+4b \\ \\ -16=4a-2b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \div 2 \\ -16=16a+4b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\ \\ -8=2a-b \\ \underline{-4=4a+b} \\ -12=6a \\ \frac{-12}{6}=a \\ a=-2 \\ \\ -8=2a-b \\ -8=2 \times (-2)-b \\ -8=-4-b \\ -8+4=-b \\ -4=-b \\ b=4$ The function is: $\boxed{y=-2x^2+4x-4}$