Finding the equation of a line with limited information.

A line with a slope of 2 is tangent to the parabola y=x^2-6x-7. Find the equation of this line. At what point does the line meet the parabola?

I'm in grade 11 university math. The unit we're doing is quadratic function. Usually, we find points of intersection by substituting, factoring, using the quadratic function or completing the square to solve. How would I go about answering this question, thanks! (No graphing calculators please!)

y = 2x + b
y = x^2 - 6x - 7
y' = 2x - 6 = 2
2x = 8
x = 4
y(4) = 16 - 24 - 7 = -15
P(4 , -15)
y + 15 = 2(x - 4)
y = 2x - 23

here is your graph:

http://www.wolframalpha.com/input/?i=y%3…

Edit: Re: Additional details:
Well, if a line is tangent to a curve at a point P(x , y) then the slope of the curve at point p is equal the slope of the line. Now the slope of the curve at a point is the derivative of the curve at that point.
In this case we were not given the point, but the slope of the line instead, so we set the derivative of the curve equal to that slope and we found the x coordinates of the point P, then we plug that into our equation and we found y coordinates -15, then P is at(4 , -15). Having a point and the slope then we found the equation of the tangent line at the point P.

Now: to solve it without using derivative:
let the line be: y = 2x + b => in the form of y = mx + b, so we have
y = 2x + b => x-intercept is -b/2 and y-intercept is b
y = x^2 - 6x - 7 => let's solve the system of equations:
y = 2x + b
y = y
x^2 - 6x - 7 = 2x + b
x^2 - 8x -(b + 7) = 0
since the line and the curve meet only once(tangent) then the quadratic has one sigle solution, using discriminant: d = b^2 - 4ac = 0 => for one solution set d to 0:
d = 64 + 4(b + 7) = 0
64 + 4b + 28 = 0
4b = -92
b = - 23
hence the equation of the line is:
y = 2x - 23

First derivative of the function y = x^2 - 6x - 7 is the slope of the tangent line at this point.