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

1) The point $(-2,1)$ is on the curve. 2) The tangent line to the curve at $(-2,1)$ is $y=-x-1$ 3) The normal line to the curve at $(-2,1)$ is $y=x+3$
$$y^2-2x-4y-1=0$$ 1) Verify that the point $(-2,1)$ is on the curve We substitute the coordinates of the point $(-2,1)$ into the formula of the function to see if it equals $0$ or not: $$1^2-2(-2)-4\times1-1=1+4-4-1=0$$ Since it equals $0$, the point is on the curve. 2) Find the tangent at $(-2,1)$: - Find the derivative of the function with implicit differentiation: $$2yy'-2-4y'-0=0$$ $$2yy'-4y'=2$$ $$yy'-2y'=1$$ $$y'(y-2)=1$$ $$y'=\frac{1}{y-2}$$ - The slope of the tangent line at $(-2,1)$ is $$y'=\frac{1}{1-2}=-1$$ - The tangent line to the given curve at $(-2,1)$ is $$y-1=-(x+2)$$ $$y-1=-x-2$$ $$y=-x-1$$ 3) Find the normal line at $(-2,1)$: The normal line at $(-2,1)$ would be perpendicular with the tangent line at $(-2,1)$, so the product of the slopes of these two lines equals $-1$. So if we call the slope of the normal line at $(-2,1)$ $k$, we would have $$k\times(-1)=-1$$ $$k=1$$ - The normal line to the given curve at $(-2,1)$ therefore is $$y-1=x+2$$ $$y=x+3$$