Exterior angles of triangles: help me find the value of x please
2 answers:
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First find the tangent line
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
![[\frac{x^3}{3}-6x^2+36x]\limits^6_0=(\frac{6^3}{3}-6(6)^2+36(6))-(0)=\frac{216}{3}-216+216=](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%2B36x%5D%5Climits%5E6_0%3D%28%5Cfrac%7B6%5E3%7D%7B3%7D-6%286%29%5E2%2B36%286%29%29-%280%29%3D%5Cfrac%7B216%7D%7B3%7D-216%2B216%3D)
the area under the curve bounded by the lines and the x axis is 72 square units
dy/dx=2x
at x=6, the slope is 2(6)=12
so
use point slope form
y-y1=m(x-x1)
point is (6,36)
so
y-36=12(x-6)
y-36=12x-72
y=12x-36
alright, so we know they intersect at x=6
and y=12x-36 is below y=x^2
so we do
the area under the curve bounded by the lines and the x axis is 72 square units