# Equation of a tangent to a curve calculus help!

• Jan 31st 2009, 01:28 PM
kimberlyann
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.

A.
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
7-3 =2x
4=2x
x =2

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
b= -20

6. The equation of the tangent is y = 7x - 20

Am I way off, or at least on the right track??? Thank you!
• Jan 31st 2009, 01:38 PM
skeeter
Quote:

Originally Posted by kimberlyann
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.

A.
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
7-3 =2x correction ... 7+3 = 2x
4=2x
x =2

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
b= -20

6. The equation of the tangent is y = 7x - 20

Am I way off, or at least on the right track??? Thank you!

way off because of that simple error in algebra at the beginning.