1. ## Tangent Line

Find an equation of the line that is tangent to the graph of f for the given value of x.

f(x)=(x^2-3)^5(2x-1)^3 x=2

I got the derivative 2(1-2x)^2(x^2-3)^4(13x^2-5x-9)

Then I plugged in 2 for x and I got 594

I'm not sure if this is right and where to go next..thanks

2. $f(x)=(x^2-3)^5(2x-1)^3... x=2$

$f'(x)=5(2x)(x^2-3)^4(2x-1)^3+3(2)(2x-1)^2(x^2-3)^5$

$f'(2)=5(4)(4-3)^4(3)^3+6(4-1)^2(4-3)^5=20(27)+54=540+54=594$

now

$f(x)=(x^2-3)^5(2x-1)^3...f(2)=27$

the equation of the tangent

$594=\left(\frac{(y-27)}{(x-2)}\right)$