posted by Jack on .
Determine the coordinates of the point (xy) where the curves y=(12)x−5 and y=x2+2x−15 intersect in the third quadrant
y = 12x - 5. y = x^2 + 2x - 15.
Substitute 12x - 5 for y in the 2nd Eq:
12x -5 = x^2 + 2x -15,
Solve for x,
12x - x^2 - 2x = -15 + 5,
-x^2 + 10x + 10 = 0,
Multiply each side by -1:
x^2 - 10x - 10 = 0,
Use quadratic formula to solve for x:
x = (10 +- sqrt(100+40)) / 2,
x = 10.92 , -0.916,
x = -0.916(3rd quadrant).
Substitute -0.916 for x in 1st Eq:
y = 12 * -0.916 - 5 = -16.
P(-0.916 , -16)
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