## Calculus 10th Edition

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