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.
!
~ EP
