To evaluate the area of the triangle ABC, we must find out first the x and y intercepts of the graph of the function f(x) = -2x + 5.
To calculate x intercept, we'll have to cancel y = 0.
But y = f(x) => f(x) = 0 <=> -2x + 5 = 0 => -2x = -5 => x = 5/2 => x = 2.5
The intercepting point of the line -2x + 5 and x axis is B(2.5 , 0).
To calculate y intercept, we'll have to put x = 0.
f(0) = -2*0 + 5
f(0) = 5
The intercepting point of the line -2x + 5 and y axis is C(0 , 5).
By definition, x axis is perpendicular to y axis,therefore the triangle 0BC is right angled triangle and the area is the half of the product of cathetus OB and OC.
A = OB*OC/2
We'll calculate the lengths of the cathetus OB and OC.
OB = sqrt[(xB-xO)^2 + (yB-yO)^2]
OB = sqrt [(2.5)^2]
OB = 2.5
OC = sqrt[(xC-xO)^2 + (yC-yO)^2]
OC = sqrt 5^2
OC = 5
A = 5*2.5/2
A = (2.5)^2
The requested area of the right angled triangle bounded by x and y axis and the graph of f(c) is: A = 6.25 square units.