# Solve Tangent Lines Problems in Calculus

Tangent lines problems and their solutions, using first derivatives, are presented.



## Problem 1

Find all points on the graph of $$y = x^3 - 3 x$$ where the tangent line is parallel to the x axis (or horizontal tangent line).
Solution to Problem 1:
Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of $$y = x^3 - 3 x$$ is given by the first derivative $$y '$$.
$$y ' = 3 x^2 - 3$$
We now find all values of $$x$$ for which $$y ' = 0$$.
$$3 x^2 - 3 = 0$$
Solve the above equation for $$x$$ to obtain the solutions.
$$x = - 1$$ and $$x = 1$$
The above values of x are the x coordinates of the points where the tangent lines are parallel to the x axis. Find the y coordinates of these points using $$y = x^3 - 3 x$$.
for $$x = - 1$$ , $$y = 2$$
for $$x = 1$$ , $$y = - 2$$
The points at which the tangent lines are parallel to the x axis are: $$(-1 , 2)$$ and $$(1 , -2)$$. See the graph of $$y = x^3 - 3 x$$ below with the tangent lines.

## Problem 2

Find the constants $$a$$ and $$b$$ so that the line $$y = - 3 x + 4$$ is tangent to the graph of $$y = a x^3 + b x$$ at $$x = 1$$.
Solution to Problem 2:
In order to find $$a$$ and $$b$$, we need to determine two algebraic equations in $$a$$ and $$b$$. The point of tangency is on the graph of $$y = a x^3 + b x$$ and on the graph of $$y = - 3 x + 4$$ at $$x = 1$$. Hence the y-coordinates of $$y = a x^3 + b x$$ and $$y = - 3 x + 4$$ at $$x = 1$$ are equal which after substitution of $$x = 1$$ gives the equation
$$a \; (1)^3 + b \; (1) = - 3(1) + 4$$
Simplify the above equation in $$a$$ and $$b$$ to obtain
$$a + b = 1$$
The slope of the tangent line is equal to $$-3$$ which is also equal to the first derivative $$y '$$ of $$y = a x^3 + b x$$ at $$x = 1$$
$$y ' = 3 a x^2 + x = - 3$$ at $$x = 1$$.
Substitute $$x = 1$$ in $$3 a x^2 + x = - 3$$ to obtain a second equation in $$a$$ and $$b$$
$$3 a + b = -3$$
Use any method to solve the system of equations $$a + b = 1$$ and $$3 a + b = - 3$$ obtained above to get the solution:
$$a = - 2$$ and $$b = 3$$.
See graphs of $$y = a x^3 + b x$$, with $$a = - 2$$ and $$b = 3$$, and $$y = - 3 x + 4$$ below.

## Problem 3

Find conditions on $$a$$ and $$b$$ so that the graph of $$y = a \; e^x + b \; x$$ has NO tangent line parallel to the x axis (horizontal tangent).
Solution to Problem 3:
The slope of a tangent line is given by the first derivative $$y '$$ of $$y = a \; e^x + b \; x$$. Hence
$$y ' = a \; e^x + b$$
To find the x coordinate of a point at which the tangent line to the graph of $$y$$ is horizontal, solve $$y ' = 0$$ for x (slope of a horizontal line is equal to zero)
$$a e^x + b = 0$$
Rewrite the above equation as follows
$$e^x = - b/a$$
The above equation has solutions for $$-a / b \gt 0$$. Hence, the graph of $$y = a \; e^x + b \; x$$ has NO horizontal tangent line if $$- a/b \le 0$$

## Exercises

1) Find all points on the graph of $$y = x^3 - 3 x$$ where the tangent line is parallel to the line whose equation is given by $$y = 9 x + 4$$.
2) Find $$a$$ and $$b$$ so that the line $$y = - 2$$ is tangent to the graph of $$y = a x^2 + b x$$ at $$x = 1$$.
3) Find conditions on $$a$$, $$b$$ and $$c$$ so that the graph of $$y = a \; x^3 + b \; x^2 + c \; x$$ has ONLY ONE tangent line parallel to the x axis (horizontal tangent line).

### Solutions to the Above Exercises

1) $$(2 , 2)$$ and $$(-2 , -2)$$
2) $$a = 2$$ and $$b = - 4$$
3) $$4 b^2 - 12 \; a \; c = 0$$