SOLUTION: Y+3=(-5/2)(x-5)
Y=(5/3)x^2+5x-12)
How do I solve the above system using an algebraic method. I also need to verify my results by graphing the system. I also have to find the area
Question 389468: Y+3=(-5/2)(x-5)
Y=(5/3)x^2+5x-12)
How do I solve the above system using an algebraic method. I also need to verify my results by graphing the system. I also have to find the area bounded by the system. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! 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 =(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.
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