# Chapter 2 - Differentiation - 2.2 Exercises: 60

There is a horizontal tangent at the point $(0, 9)$.

#### Work Step by Step

Using the power rule $((x^n)'=nx^{n-1})$ we get $y'=2x$ Horizontal tangent$\rightarrow y'=0 \rightarrow 2x=0 \rightarrow x=0$ Hence the derivative is $0$ when $x=0$; to find the y-coordinate substitute in the original equation to get $y=(0)^2+9\rightarrow y=9\rightarrow$The point is $(0, 9)$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.