Posted by **Jack** on Wednesday, September 8, 2010 at 9:33pm.

Determine the coordinates of the point (xy) where the curves y=(12)x−5 and y=x2+2x−15 intersect in the third quadrant

- Pre Calculus -
**Henry**, Thursday, September 9, 2010 at 12:02pm
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,

Combine like-terms:

-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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