Answer:
![f(x) =2x^(2) +4x-7](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k2njtqa5f33cxjrym4obyzmli52j4ke9t0.png)
Explanation
Let equation be
![f(x) =ax^(2) +bx+c](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ykd2b5cwk8gu5ljtditng78tdceiu9fljn.png)
Since (-4,9) satisfy the equation ,plugging x =-4 and f(x) = 9
![f(x) =a(-4)^(2) +b(-4)+c](https://img.qammunity.org/2020/formulas/mathematics/middle-school/z25m83pt76oygpdqitxv2jshwik2p8l6ql.png)
9 = 16a -4b +c ........ equation 1
plugging x =0 and f(x) = -7
-7 = a(0) +b(0) +c
gives c =-7 ......... equation 2
plugging x =1 and f(x) =-1
-1 = a(1) + b(1) +c
-1 = a+b+c ............. equation 3
From equation 2 ,we have c =-7
Plugging this value in equation in 1 and 3 ,we get
16a -4b -7 = 9
16a-4b= 9+7
16a-4b = 16
Dividing the equation by 4 ,we get
4a -b = 4 .......... equation 4
Plugging c =-7 in equation 1 ,we get
a+b+(-7) = -1
a+b = -1+7
a+b = 6 ......... equation5
Adding equation 4 and equation 5
a+b =6
4a-b =4
______________
5a = 10
a = 2
plugging a =2 in equation2
2+b = 6
b = 6-2
b = 4
therefore quadratic equation is
![f(x) =2x^(2) +4x-7](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k2njtqa5f33cxjrym4obyzmli52j4ke9t0.png)