Answer to Question #252563 in Calculus for Mary

Question

Find the area of the triangle formed from the coordinate axes and the tangent line to the curve y = 5x^(−1) −x/5 at the point (5,0).

Answer 1

$y=\dfrac{5}{x}-\dfrac{x}{5}$

Find the first derivative

$y'=-\dfrac{5}{x^2}-\dfrac{1}{5}$

Point $(5,0)$

$slope=m=y'(5)=-\dfrac{5}{5^2}-\dfrac{1}{5}=-\dfrac{2}{5}$

The equation of the tangent to the curve at the point (5, 0) in point-slope form

$y-0=-\dfrac{2}{5}(x-5)$

The equation of the tangent to the curve at the point (5, 0) in slope-intercept form

$y=-\dfrac{2}{5}x+2$

$y$-intercept: $x=0=>y=-\dfrac{2}{5}(0)+2=2.$ Point $(0, 2).$

$x$-intercept: $0=0=>y=-\dfrac{2}{5}x+2=>x=5.$ Point $(5, 0).$

The area of the right triangle is equal to

$A=\dfrac{1}{2}(2)(5)=5({units}^2)$

The area of the triangle formed from the coordinate axes and the tangent line to the curve $y=5x^{-1}-\dfrac{x}{5}$ at the point (5,0) is 5 squared units.