Find equations of both lines through the point (2,-3) that are tangent to the parabola y=x2+x.
Slope formula: (Y2-Y1)/(X2-X1) = M
The Attempt at a Solution
Here's what I think I need to do. First I think I need to find the derivative of the parabola. Then I need to use the point (2,-3) as point 2 in the slope equation Y2-Y1/X2-X1 to find two X values that give me a slope equal to the derivative at that x-value.
Finding slope: The tangent line passes through (2,-3) and the parabola. Since the slope of the tangent line is found by the derivative, that means that y2-y1/x2-x1 should be equal to the derivative 2x+1, or: [-3-(2x+1)]/(2-x) = 2x+1
So that becomes: (-3-2x-1)/(2-x) = 2x+1, then (-4-2x)/(2-x) = 2x+1.
Moving the bottom over: -4-2x = (2x+1)(2-x)
Foiling: -4-2x = 4x-2x2+2-x.
Combining: -4-2x = -2x2+3x+2
Moving the right side over and combining: 2x2-5x-6 = 0
So the tangent lines should be located at the x-coordinates found by finding x in the above quadratic, right?