# How do you find the points where the graph of the function f(x) = x^4-4x+5 has horizontal tangents and what is the equation?

Feb 28, 2016

at $\left(1 , 2\right)$; equation of $y = 2$

#### Explanation:

A horizontal tangent occurs whenever the function's derivative equals $0$, since a value of $0$ represents that the function's tangent line has a slope of $0$. Lines with slope $0$ are horizontal.

To find the function's derivative, use the power rule.

$f \left(x\right) = {x}^{4} - 4 x + 5$

$f ' \left(x\right) = 4 {x}^{3} - 4$

Find the points when $f ' \left(x\right) = 0$.

$4 {x}^{3} - 4 = 0$

$4 {x}^{3} = 4$

${x}^{3} = 1$

$x = 1$

There is a horizontal tangent at $\left(1 , 2\right)$, thus its equation is $y = 2$.

We can check a graph of $f \left(x\right)$:

graph{(x^4-4x+5-y)(y-0x-2)=0 [-19.92, 20.63, -3.52, 16.74]}