We have been given the expression to be [tex] y=(x-3)^7(x+1)^2 [/tex]
Since we need to find the tangent at a point, we will have to find the derivative of [tex] y [/tex] as the slope of the tangent at a given point on the curve is always equal to value of the derivative at that point.
Thus, we have to find [tex] \frac{dy}{dx}=\frac{d}{dx} (x-3)^7(x+1)^2 [/tex]
We will use the product rule of derivatives to find [tex] \frac{dy}{dx} [/tex]
Thus, [tex] y'=7(x-3)^6(x+1)^2+(x-3)^7\times2(x+1) [/tex] (using the product rule which states that [tex] (fg)'=f'g+fg' [/tex])
Taking the common factors out we get:
[tex] y'=(x-3)^6(x+1)(7(x+1)+2(x-3)) [/tex]
[tex] y'=(x-3)^6(x+1)(7x+7+2x-6)=(x-3)^6(x+1)(9x+1) [/tex]
Thus, [tex] y' [/tex] at [tex] x=4 [/tex] is given by:
[tex] y'_{x=4} [/tex]=Slope of the tangent of y at x=4=[tex] m [/tex]
Thus, [tex] m=(4-3)^6(4+1)(9\times4+1)=185 [/tex]
Now, the equation of the tangent line which passes through [tex] (x_{1}, y_{1}) [/tex] and has slope m is given by:
[tex] y-y_{1}=m(x-x_{1}) [/tex]
Thus, the equation of the tangent line which passes through [tex] (4,25) [/tex] and has the slope 185 is[tex] y-25=185(x-4) [/tex]
Which can be simplified to [tex] y=185x-740+25=185x-715 [/tex]
Thus, [tex] y=185x-715 [/tex]
This is the required equation of the tangent.