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If we graph the function f(x)=((x-2)^2-9)/3, the x and y intercepts of the graph are connected to make a polygon. What is the area?

 Oct 24, 2020
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If we graph the function f(x)=((x-2)^2-9)/3, the x and y intercepts of the graph are connected to make a polygon. What is the area?

 

Hello Guest!

 

\(f(x)=((x-2)^2-9)/3\\ =\frac{1}{3}(x^2-4x+4-9)\\ =\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}\)

 

\(f(x)=\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}=0\)

\(x^2-4x-5=0\\ x=2\pm \sqrt{4+5}\)

\(x_1=-1\\ x_2=5\\ f(0)=-\frac{5}{3}\)

The polygon is a triangle ABC.

A (-1,0)

B (5,0)

C (0, \(-\frac{5}{3}\) )

\(\triangle ABC\\ c=5-(-1)\\ \color{blue}c=6\\ h=|-\frac{5}{3}|\\ \color{blue}h=\frac{5}{3}\)

\(A=\frac{c\cdot h}{2}=\frac{6\cdot 5}{3\cdot 2}\)

\(A=5\)

 

The area of the triangle ABC is 5.

laugh  !

 Oct 25, 2020
edited by asinus  Oct 25, 2020
edited by asinus  Oct 25, 2020
edited by asinus  Oct 25, 2020

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