Find the slope of the tangent line to the graph of the given function at the given value of x. Find the...

Find the equation of the tangent line. Please do it step by step.

1. y= -3x^5-8x^3+4x^2; x=1

2. y= -x^-3+x^-2; x=2

1. first of all we need to find the value of y with the given value of x=1

y=-3*1^5 -8*1^3 +4*1^2 =-7

so the point is (1,-7)

for the slope use derivative of the function

y ' = -3*5x^4 -8*3x^2 +4*2x
y '(1)= -15 -24 +8 = -31

y -y1 = m(x-x1)

y+7 =-31(x-1)

y=-31x +24


2. find value of y by given value of x=2

y = -2^-3 +2^-2 = 1/8

so the point is (2,1/8)

for the slope use derivative of the function:

y ' = 3/x^4 -2/x^3
y '(2)= 3/16 -2/8 = -1/16

y-1/8 = -1/16(x-2)

y=(-1/16)x +1/4