Determine the equation(s) of the lines that are tangent to $y = x^2 + 3x + 1$ and pass through the origin.

Question

I am trying to solve the question but I am not sure how to approach it. The derivative is 2x+3 but does that have any significance when solving this?

Answer 1

Hint: The equation of a line tangent to a function $f(x)$ at point $x=c$ is

$$y-f(c)=f'(c)(x-c).$$

Determine all $c$ such that the line contains the origin:

$$f(c)=f'(c)c.$$

Answer 2

This one is perhaps simpler without calculus: as lines through the origin have form $y=mx$, you are looking for values of $m$ that give exactly one real root for $mx=x^2+3x+1$. So set the discriminant to zero...

Answer 3

We can easily get the tangent points of interest. We have

• $f(x)=x^2+3x+1$ as well as
• $f'(x)=2x+3$

We'll define

$g(x)=3x+2$ and set this to equal $f(x)$. So,

$x^2+3x+1=3x+2$ or simply $x^2=1$.

This gives us the x-values of our function $f(x)$ where tangents pass throught $(0,0)$. We get $x_1=-1$ and $x_2=1$, so we have the tangent points in $f(-1)$ and $f(1)$ being $t_1=(-1,-1)$ and $t_2=(1,5)$.

Thus, the two tangents are of form

$t: y_1=x$ and $t: y_2=5x$.