Find the equation(s) of the tangent line(s) to the graph of the curve y = x^3 − 10x through the point (1, −10) not on the graph. (Enter your answers as a comma-separated list of equations.)

Answer 1

Every point on the curve has the format (x, x^3-10x), we are given another point (1,-10) The slope of the curve is y'(x) = 3x^2 - 10 and that slope of the tangent line is [delta y / delta x] = (-10 - x^3 + 10x)/(1-x) and we know that the slope of the line through those two points, one on the line and one not, (-10 - x^3 + 10x)/(1-x), must be equal to 3x^2 - 10.

Let's see if there are particular x's that make this work.

(-10 - x^3 + 10x)/(1-x) = 3x^2 - 10 =>

x^3 - 10x +10 = (3x^2 -10)(x-1) = 3x^3 - 10x - 3x^2 +10

or 2x^3 - 3 x^2 = 0

x^2(2x - 3) = 0

So our conditions work when x is 0 and 3/2, in those cases we can evaluate y at each of those values and we find that the other points on the curve where the lines through (1,-10) are tangent are (0,0) and (3/2,-53/8).

From there we can calculate two linear equations that would go through each of the two pairs of points, one on the line, one not on the line. [y = -10x, and y = -(13/4)x - (27/4)]

This is the methodology; you should check for yourself that the lines go through the points and are tangent to the y curve in the right places! Also you should be sure that you understand the steps and can work out the equations from the points to the lines by yourself.

Answer 2

Let a = the x coordinate of the point on the graph that will also be the point(s) of tangency. Its y coordinate will be a³ - 10a.

The slope of the tangent line(s) at the point of tangency can be found by substituting "a" into y', i.e. 3a² - 10.

But you can also calculate the slope of the line through (1, -10) and (a, a³ - 10a) by using the formula for slope. And that expression must be equal to 3a² - 10.

So, [ a³ - 10a - (-10)]/ (a - 1) = 3a² - 10 (left hand side is change in y over change in x between the two points).

Solve that equation for "a" (getting two values), determine the two points of tangency, use the derivative to find the slopes of the tangent lines at those two points, then use point-slope to find the equations of the tangent lines passing through (1, -10).

Confirm your answer by checking out this graph:

Append the following to the link for the Desmos online graphing caculator: /dmcwlfzhxx