plug this value into the original equation, and find yy = 4^2 - 6*4 - 7 = - 15The equation of the line will be y - y1 = m (x - x1)y - (-15) = 2 (x - 4)2x - y - 23 = 0-you take the derivative of the parabolay = 2x-6now set y = 2 (slope of the line is 2)2 = 2x-6x=4plug into parabola equation to find y coordinatey=4^2 - 6(4) - 7 = -15now you have a point (4 , -15) and the slope m=2plug into point-slope formulay+15 = 2(x-4)y+15 = 2x-8y=2x - 23-Heres one approach (of several):A line with a slope of two will have the formula y = 2x + kIf its a tangent, it will touch the parabola at one point only. Can you take it from here? (Ill check back later.......
Then,
y' = 2x - 6 = 2
x = 4
Now, plug this value into the original equation, and find y
y = 4^2 - 6*4 - 7 = - 15
The equation of the line will be y - y1 = m (x - x1)
y - (-15) = 2 (x - 4)
2x - y - 23 = 0
