Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 53

Given:
Take the derivative of this function.
y' = 4x^3-6x
Substitute the value of x=1 from the given point into the derivative function.
y' = 4(1)^3-6(1)
y'=4-6 = -2
The slope at the given point is negative 2.
a)
y'=-2
c)
Graph the derivative function :

The derivative at x=1 is negative 2.
b) Our slope is -2, and we will use the slope-intercept form with the given point to find the y-intercept, and write our equation.
y=mx+b
Substitute the point and slope.
0 = (-2)(1)+b
0=-2+b
b=2
The equation of the tangent line is then:
y=-2x+2
The original function and the tangent line are graphed in the imaged attached.