The equation of the tangent line to y=f(x) at a=2 can be written in the form y=mx+b where f(x) =4sinx/4sinx+6cosx

i already got 3/2 for m but i cant seem to find out how to get the value of b?

Well, let me help you out with that! To find the value of b in the equation y = mx + b, you need to substitute the known values of a and m into the equation and solve for b.

In this case, you have already determined the slope to be m = 3/2, so the equation of the tangent line can be written as:

y = (3/2)x + b

Now, to find the value of b, we need to substitute the x-coordinate of the point of tangency (a=2) and the y-coordinate of that same point (which we still need to find) into the equation.

To find the y-coordinate, let's substitute x = 2 into the original function f(x):

f(2) = 4sin(2)/(4sin(2) + 6cos(2))

Now, calculating the function value f(2) will give you a y-coordinate you can use to solve for b in the equation y = (3/2)x + b.

Keep in mind that this is a serious mathematical calculation, so it might take more time and effort than I usually put into humor. But don't worry, I'll still be here to entertain you while you work on it!

To find the value of b, we need to use the point-slope form of a line equation. Given that the tangent line passes through the point (a, f(a)), which in this case is (2, f(2)), we can substitute these values into the equation.

First, let's find the value of f(2) by substituting x = 2 into the function f(x):

f(2) = 4sin(2) / (4sin(2) + 6cos(2))

Now, let's calculate f(2):

f(2) = 4(0.9093) / (4(0.9093) + 6(0.383))

f(2) ≈ 3.6367 / (3.6372 + 2.298)

f(2) ≈ 3.6367 / 5.9352

f(2) ≈ 0.6127

Therefore, the point (2, f(2)) is approximately (2, 0.6127).

Now, we can use the point-slope form of a line equation (y - y1) = m(x - x1), where (x1, y1) is the point (2, 0.6127) and m = 3/2:

(y - 0.6127) = (3/2)(x - 2)

Rearrange the equation to get it in slope-intercept form (y = mx + b):

y - 0.6127 = (3/2)(x - 2)

y - 0.6127 = (3/2)x - 3

y = (3/2)x - 3 + 0.6127

y = (3/2)x - 2.3873

Therefore, the equation of the tangent line to y = f(x) at a = 2 can be written as y = (3/2)x - 2.3873, where m = 3/2 and b = -2.3873.

To find the value of b, which represents the y-intercept of the tangent line, we need to use the given equation of the function f(x).

The equation of the tangent line can be found using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point of tangency.

In this case, the point of tangency is (a, f(a)), where a = 2. So, the point becomes (2, f(2)).

To find f(2), substitute x = 2 into the given equation for f(x):

f(x) = 4sinx / (4sinx + 6cosx)
f(2) = 4sin2 / (4sin2 + 6cos2)

Now, calculate f(2) using this equation.

Then substitute the calculated value of f(2) and the value of m, which you've already found to be 3/2, into the point-slope form:

y - f(2) = (3/2)(x - 2)

Now, rearrange the equation to isolate y:

y = (3/2)(x - 2) + f(2)

And there you have it! The equation of the tangent line in the form y = mx + b, where m = 3/2 and b = f(2).

Does f(x) = 4 sinx/(4sinx+6cosx) ?

If so, you need parentheses in the denominator.
You say you want the tangent point to be at a=2. Do you mean x=2? Or x = 2 pi? I don't see how you can get 3/2 for the slope if the equation involves the sin and cos of 2.

If you have the correct slope m, then you can get b by requiring that the value of y = mx + b be the same at the value of f(x) at at the point of contact.