Question 389468
y+3 = (-5/2)(x-5)
Y =(-5/2)x+(25/2)-3
   =(-5/2)x+(25/2)-(6/2)
   =(-5/2)x+(19/2)
This function is a straight line with a slope of -5/2 and y-intercept of 19/2

second function:
y =(5/3)x^2+5x-12
This function is a parabola that opens upwards and has a y-intercept of -12

To solve, set both equations equal to each other:

(-5/2)x+(19/2) =(5/3)x^2+5x-12
multiply both sides of the equation by the LCD=6
-15x+57 =10x^2+30x-72
10x^2+45x-129 = 0
use following quadratic equation to solve with a=10, b=45, and c=-72
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

x =(45±84.76)/20 = -6.49 or 1.99
y = (-5/2)* (-6.49)+(19/2) =25.7
y = (-5/2)* (1.99)+(19/2) = 4.53

Points of intersection are (-6.49,25.7) and (1.99,4.53)

The area bounded is the bottom of the parabola cut off by the straight line at the points of intersection

You can confirm these figures by using a graphing calculator like a TI-83 or a PC graphing program which I have done.


{{{ graph( 300, 200, -6, 5, -10, 10, -2.5x+9.5,1.67x^2+5x-12) }}}