instantaneous rate of change

Question

Find the slope of the line tangent to the graph of the function at the given value of x. y=x^4+2x^3+2x+2 at x=-3

Answer 1

The slope of the tangent line at x = -3

is simply m= f^' (-3) =4(-3)³+6(-3)² +2 = -52

Answer 2

y(-3) = (-3)^4 + 2(-3)^3 +2(-3) + 2

= 81 -54 -6 +2 = 23 = y when x= -3 it's the point (-3, 23)

y' = 4x^3 + 6x^2 + 2 = slope of the tangent line = the derivative of y(x) = y'(x)

y'(-3) = 4(-3)^3 +6(-3)^2 + 2 = -108 + 54 + 2 = -52 = m = slope of tangent line at x=-3

y=mx + b where m=slope and b = y intercept

solve for b by plugging in the point (-3, 23) and m= -52

23=-52(-3) + b

b= 23- 156= -133

y=-52x -133 is the tangent line at x=-3 at the point (-3, 23)