Answer to Question #252563 in Calculus for Mary

Question #252563

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).


1
Expert's answer
2021-10-20T02:21:42-0400
"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.



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