Find the equation(s) of the tangent line(s) to the graph of the curve y = x3 − 11x through the point (1, −11) not on the graph. (Enter your answers as a comma-separated list.)

Answer 1

The derivative is 3x^2 - 11 at (x, x^3 - 11x)

Thus, the equation of the tangent is y = (3x^2 - 11)x + b;

solving for b, we get,

x^3 - 11x = (3x^2 - 11)x + b, or

x^3 - 11x = 3x*3 - 11x + b, so b = -2x^2

y = (3x^2 - 11)x - 2x^2

Then, as it also goes through (1, -11),

Thus, -11 =(3x^2 - 11)1 -2x^2

Thus, 0 = x^2

x = 0

The equation of the tangent line is

y = -11x

As a check, for the curve, f(0) = 0^3 - 11*0 = 0

f'(0) = 3*0^2 = -11

y = mx + b

0 = -11(0) + b

b = 0

y = -11x is the tangent line

At x= 1, y = -11(1) = -11, so the tangent goes through (1, -11)

Answer 2

The slope of any tangent line is 3x²-11.

Therefore the equation of any tangent line through (1,-11) is 3x²-11=(y+11)/(x-1)

Substituting x³-11x for y in the above equation gives (after some algebraic manipulation)

2x³-3x²=0 which means that the points of tangency occur when x=0 and x=3/2.

Since x=0 is a double root, there are only 2 tangents:

Substitute each of the values of x into the equation for the derivative and then use the point slope equation::

-11=(y+11)/(x-1) and -17/4=(y+11)/(x-1).

I will leave it to you to "massage" these equations into any necessary form.

However, I will tell you that I went to DESMOS and graphed these results and they appear to be correct.