# How do you find the equation of a line tangent to the function #y=sqrt(2x+1)# at x=4?

The tangent has equation

Start by finding the corresponding y-coordinate.

Find the derivative.

Find the slope of the tangent by evaluating the point within the derivative.

The equation of the tangent is therefore:

Hopefully this helps!

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To find the equation of a line tangent to the function y = √(2x + 1) at x = 4, we need to determine the slope of the tangent line and the point of tangency.

First, we find the derivative of the function y = √(2x + 1) with respect to x.

dy/dx = (1/2) * (2x + 1)^(-1/2)

Next, we substitute x = 4 into the derivative to find the slope of the tangent line at x = 4.

dy/dx = (1/2) * (2(4) + 1)^(-1/2) = (1/2) * (9)^(-1/2) = 1/6

So, the slope of the tangent line at x = 4 is 1/6.

To find the point of tangency, we substitute x = 4 into the original function y = √(2x + 1).

y = √(2(4) + 1) = √(9) = 3

Therefore, the point of tangency is (4, 3).

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line, we can substitute the values to find the equation of the tangent line.

y - 3 = (1/6)(x - 4)

Simplifying the equation, we get:

y = (1/6)x + 2/3

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