# How do you find the equation of the line that is tangent to f(x)=x^3 and parallel to the line 3x-y+1=0?

##### 1 Answer
Oct 26, 2016

$y = 3 x \pm 2$

#### Explanation:

The tangent in ${x}_{0}$ must have $m = {f}^{'} \left({x}_{0}\right) = 3 {x}_{0}^{2}$ and at the same time be parallel to $y = 3 x + 1 \implies m = 3$
so
${x}_{0}^{2} = 1$
and
${x}_{0} = \pm 1 \implies {y}_{0} = \pm 1$
the tangent equation is
$\left(y - {y}_{0}\right) = 3 \left(x - {x}_{0}\right)$
$\left(y \pm 1\right) = 3 \left(x \pm 1\right)$
$y = 3 x \pm 2$