Question

Let f(x)=sin(x) + 2 cos(x). Find the slope-intercept equation of the line tangent to the graph of y = f(x) at the point on the graph where x = 0. The equation of the tangent line is (write your answer in the form y =mx+b):​

Answer 1

The equation of the tangent line at x = 0 can be found by differentiating the function f(x) and using the point-slope form of a line.

The derivative of f(x) is f'(x) = cos(x) - 2sin(x). Substituting x = 0 into f'(x) gives f'(0) = cos(0) - 2sin(0) = 1 - 0 = 1.

Therefore, the equation of the tangent line is y = 1x + b, or y = x + b.

To find the value of b, we can plug in x = 0 and y = f(0) = sin(0) + 2cos(0) = 2 into the equation. Solving for b, we get 2 = 0 + b, so b = 2.

Therefore, the slope-intercept equation of the line tangent to the graph of y = f(x) at the point (0,2) is y = x + 2.

Answer 2

Explanation:

The first step is to find the slope of the tangent line. To do that, we need to find the derivative of f(x):

f(x) = sin(x) + 2cos(x)

f'(x) = cos(x) - 2sin(x)

Now we can find the slope of the tangent line at x=0 by plugging in x=0 into f'(x):

f'(0) = cos(0) - 2sin(0) = 1

Therefore, the slope of the tangent line at x=0 is 1.

Next, we need to find the y-coordinate of the point on the graph where x=0. To do that, we simply plug in x=0 into f(x):

f(0) = sin(0) + 2cos(0) = 2

Therefore, the point on the graph where x=0 is (0, 2).

Now we can use the point-slope form of the equation of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is the point on the line. Plugging in m=1 and (x1, y1) = (0, 2), we get:

y - 2 = 1(x - 0)

Simplifying, we get:

y = x + 2

Therefore, the equation of the tangent line is y = x + 2.