# Write the equation of the tangent line to the graph of the function at the indicated point. Check the reasonableness of your answer by graphing both the function and the tangent line.

Y=(x-2)/(15-x^2)

x=-4

HOW DO WE DO THIS

## its okay i got it

## Interesting - you figured it out in 9 minutes. Maybe help here is so good that just *asking* for help helps yo solve the problem? :-)

## To find the equation of the tangent line to the graph of a function at a given point, we can follow these steps:

1. Find the derivative of the function.

2. Evaluate the derivative at the given x-coordinate to find the slope of the tangent line.

3. Use the point-slope form of a linear equation to form the equation of the tangent line using the slope and the given point.

Now let's apply these steps to the specific problem:

1. Find the derivative of the function Y = (x-2)/(15-x^2):

- To simplify the calculation, we can rewrite the function using a common denominator:

Y = (x-2)/[(sqrt(15)-x)(sqrt(15)+x)].

- Apply the quotient rule to find the derivative:

Y' = [((sqrt(15)-x)(sqrt(15)+x)) - (x-2)(2x)]/[(sqrt(15)-x)(sqrt(15)+x)]^2.

- Simplify the expression inside the brackets:

Y' = [(15 - x^2 + 2x - 2x^2) - (2x^2 - 4x)]/[(sqrt(15)-x)(sqrt(15)+x)]^2.

Y' = (15 - 3x^2)/[(sqrt(15)-x)(sqrt(15)+x)]^2.

- Simplify further if desired.

2. Evaluate the derivative at the given x-coordinate, x = -4:

Y'(-4) = (15 - 3(-4)^2)/[(sqrt(15)-(-4))(sqrt(15)+(-4))]^2.

Y'(-4) = (15 - 3(16))/[(sqrt(15)+4)(sqrt(15)-4)]^2.

Y'(-4) = (15 - 48)/[(sqrt(15)+4)(sqrt(15)-4)]^2.

3. Use the point-slope form of a linear equation and the given point (-4, Y(-4)) = (-4, Y(-4)) to write the equation of the tangent line:

Y - Y(-4) = Y'(-4)(x - (-4)).

Y - Y(-4) = Y'(-4)(x + 4).

Now you have an equation of the tangent line in point-slope form. To check the reasonableness of the answer, you can graph both the original function and the tangent line.