In order to find the equation of the tangent line to the curve at P(3,-2) first we would need to find the derivative of the function.
In this case y=x^2 -3x-2
Hence, the derivative would be:
f'(y)=2x -3
So, from the point (3,-2) we can obtain the x value which is 3.
Therefore, we would substitute the number 3 on f'(y)=2x -3.
So, f'(3)=2(3)-3
f'(3)=6-3=3
So, to form the equation we would have to base it from the original equation:
y=mx +b
m=3, because f'(3)=3.
So, we would use the P(3,-2) to give values to x and y and find b.
So:
-2=3(3)+b
-2=9+b
b=-11
if b=-11, hence, the equation of the tangent line to the curve at P(3,-2) would be:
y=3x -11