# Thread: Equation of Line tangent to Curve.

1. ## Equation of Line tangent to Curve.

Find an equation of the tangent line to the curve at the point (0, 0).

$y=sin(6 x) + sin^2(6 x)$

I found the derivative to be $6cos(6x)(1+2sin(6x))$.

When I plug in 0 for x I get y = 6 but I am not sure how that helps me. Galactus explained this to me the other night but I looked over that thread and can't apply what he showed me to this problem.

2. Remember that when you find a derivative, you're finding a formula that will give you the slope of the tangent line to any point on your function. So, when you plug in x = 0 into your derivative and get out y = 0 (good!), that value y = 0 is in SLOPE of the tangent line to your function at the point (0,0).

So now you now your tangent line has a slope of 0. Think about what kinds of lines have a slope of 0 - this should give you an idea of what your line is. Remember, lastly, that your line has to pass through your original point, (0,0). So what is the equation of a line with a slope of 0 that passes through (0,0)?

3. But when I plugged 0 for x into the derivative I got 6.

4. Topher,

Sorry - should've proofread. Here's an updated response:

Remember that when you find a derivative, you're finding a formula that will give you the slope of the tangent line to any point on your function. So, when you plug in x = 0 into your derivative and get out y = 6 (good!), that value y = 6 is in SLOPE of the tangent line to your function at the point (0,0).

So now you now your tangent line has a slope of 6. You also know that it goes through (0,0). So you need the equation of a line given a point and the slope. This suggests using point-slope form to write your equation.

Point-slope form, in case you've forgotten, is:

(y - y_1) = m (x - x_1).

Your slope is m and your point, (0,0), plays the role of (x_1, y_1). Plug in those three things and you should be in good shape.

Sorry about the confusion! Too many zeros floating around, I guess.