Equation of a tangent to a curve calculus help!
So for the following question, I have gone through the steps to solve it. I am just not confident in my answer so if anyone could help that would be appreciated!
Q. Find the equation of the tangent to y = x^2 - 3x - 4 that is parallel to the line y = 7x + 3.
1. The slope of y=7x+3 is 7, hence the slope of the tangent is 7.
2. Finding the point on the curve where the derivative is 7.
y'= 2x - 3
7= 2x -3
3. Solving for y;
y= (2)^2 - 3(2) - 4
y = -6
4. The point on the curve is (2, -6)
5. Using the coordinates to find the y-intercept:
y= 7x + b
-6 = 7(2) + b
6. The equation of the tangent is y = 7x - 20
Am I way off, or at least on the right track??? Thank you!