## University Calculus: Early Transcendentals (3rd Edition)

1) The point $(-1,0)$ is on the curve. 2) The tangent line to the curve at $(-1,0)$ is $$y=\frac{6}{7}x+\frac{6}{7}$$ 3) The normal line to the curve at $(-1,0)$ is $$y=-\frac{7}{6}x-\frac{7}{6}$$
$$6x^2+3xy+2y^2+17y-6=0$$ 1) Verify that the point $(-1,0)$ is on the curve We substitute the coordinates of the point $(-1,0)$ into the formula of the function to see if it equals $0$ or not: $$6\times(-1)^2+3\times(-1)\times0+2\times0^2+17\times0-6$$ $$=6+0+0+0-6=0$$ Since it equals $0$, the point is on the curve. 2) Find the tangent at $(-1,0)$: - Find the derivative of the function with implicit differentiation: $$12x+3(y+xy')+2yy'+17y'-0=0$$ $$12x+3y+3xy'+2yy'+17y'=0$$ $$3xy'+2yy'+17y'=-12x-3y$$ $$y'(3x+2y+17)=-(12x+3y)$$ $$y'=-\frac{12x+3y}{3x+2y+17}$$ - The slope of the tangent line at $(-1,0)$ is $$y'=-\frac{12\times(-1)+3\times0}{3\times(-1)+2\times0+17}=-\frac{-12}{-3+17}=-\frac{-12}{14}=\frac{6}{7}$$ - The tangent line to the given curve at $(-1,0)$ is $$y-0=\frac{6}{7}(x+1)$$ $$y=\frac{6}{7}x+\frac{6}{7}$$ 3) Find the normal line at $(-1,0)$: The normal line at $(-1,0)$ would be perpendicular with the tangent line at $(-1,0)$, so the product of the slopes of these two lines equals $-1$. So if we call the slope of the normal line at $(-1,0)$ $k$, we would have $$k\times\frac{6}{7}=-1$$ $$k=-\frac{7}{6}$$ - The normal line to the given curve at $(-1,0)$ therefore is $$y-0=-\frac{7}{6}(x+1)$$ $$y=-\frac{7}{6}-\frac{7}{6}$$