Answer to Question #120815 in Calculus for Aryan Patel

Question

Find the equations of tangent and normal to the curve y=4x-x^2 at (1,3)

Answer 1

Equations of tangent line and normal line can be found using:

$y-y_0=y'(x_0) (x-x_0) -$ tangent line

$y-y_0=-\frac{1}{y'(x_0)}(x-x_0) -$ normal line

So, all we need is to find $y'(x_0)$

$x_0=1\\y_0=3$

$y'(x)=(4x-x^2) '=4-2x$, then

$y'(x_0) =y'(1) =4-2*1=2$

Finally:

$y-3=2(x-1) \iff y=2x+1$ — tangent line

$y-3=-\frac{1}{2}(x-1) \iff y=-\frac{1}{2}x+\frac{7}{2}$ — normal line