We can apply some rules backwards
first
f'(x)=
(x-4)(-2x+4)
-2x^2+12x-16
we know that
![f'(rx^n)=rnx^{n-1}](https://tex.z-dn.net/?f=f%27%28rx%5En%29%3Drnx%5E%7Bn-1%7D)
so therefor maybe
-2x^2=rnx^{n-1}
2=n-1
n=3
rn=-2
3r=-2
r=-2/3
one is
![\frac{-2}{3}x^3](https://tex.z-dn.net/?f=%20%5Cfrac%7B-2%7D%7B3%7Dx%5E3%20)
second part
12x
12x=
![rnx^{n-1}](https://tex.z-dn.net/?f=rnx%5E%7Bn-1%7D)
x^1, 1=n-1
n=2
rn=12
2r=12
r=6
![6x^2](https://tex.z-dn.net/?f=6x%5E2)
is the second bit
last part
-16
-16x^0=
![rnx^{n-1}](https://tex.z-dn.net/?f=rnx%5E%7Bn-1%7D)
0=n-1
n=1
rn=-16
1r=-16
r=-16
-16x^1
so therfor f(x)=
For this case we have that by definition, the equation of the line in the slope-intersection form is given by:
![y = mx + b](https://tex.z-dn.net/?f=y%20%3D%20mx%20%2B%20b)
Where:
m: It is the slope of the line
b: It is the cut-off point with the y axis
According to the statement, the line goes through the following points:
![(x_ {1}, y_ {1}) :( 1,1)\\(x_ {2}, y_ {2}) :( 0, -3)](https://tex.z-dn.net/?f=%28x_%20%7B1%7D%2C%20y_%20%7B1%7D%29%20%3A%28%201%2C1%29%5C%5C%28x_%20%7B2%7D%2C%20y_%20%7B2%7D%29%20%3A%28%200%2C%20-3%29)
We found the slope:
![m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {-3-1} {0-1} = \frac {-4} {- 1 } = 4](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%20%7By_%20%7B2%7D%20-y_%20%7B1%7D%7D%20%7Bx_%20%7B2%7D%20-x_%20%7B1%7D%7D%20%3D%20%5Cfrac%20%7B-3-1%7D%20%7B0-1%7D%20%3D%20%5Cfrac%20%7B-4%7D%20%7B-%201%20%7D%20%3D%204)
Thus, the equation is of the form:
![y = 4x + b](https://tex.z-dn.net/?f=y%20%3D%204x%20%2B%20b)
We substitute one of the points and find b:
![-3 = 4 (0) + b\\b = -3](https://tex.z-dn.net/?f=-3%20%3D%204%20%280%29%20%2B%20b%5C%5Cb%20%3D%20-3)
Thus, the equation is of the form:
![y = 4x-3](https://tex.z-dn.net/?f=y%20%3D%204x-3)
Answer:
![y = 4x-3](https://tex.z-dn.net/?f=y%20%3D%204x-3)