Find the y-intercept of the tangent to y=2/x+1 at (1,1)

Question

Find the y-intercept of the tangent to y=2/x+1 at (1,1)

Answer 1

For any problem in which it asks you to find the tangent of a function at a point on that function we use...

Point Slope Form:

y-y₁=m(x-x₁)

In which x₁ and y₁ is from your given point.

And m is found by taking the derivative of the original equation at the point given, f'(x₁).

In this case a tangent to the function y=2/(x+1) at the point (1,1). Note (1,1)=(x₁,y₁)

So f'(x) = (d/dx)(2/(x+1)) = (-2/(x+1)^2)

f'(x₁) = f'(1) = (-2/(1+1)^2) = -1/2 (m value!)

[y-y₁=m(x-x₁)] -> [y-1=(-1/2)(x-1)] -> [y=(-1/2)(x-3)] This is the tangent to the function at the given point.

To find this tangent's y-intercept, aka where this tangent intersects the y-axis, plug in 0 for x in the tangent function.

[y(0)=(-1/2)(0-3)] -> (0,1.5) or y=1.5

Therefore, your answer is y=1.5

Answer 2

y' = -2(x + 1)^-2

At x = 1, slope of tangent is -1/2

Use point slope to determine equation of tangent and then y-intercept.