# Math Help - Find the Tangent Line Equation

1. ## Find the Tangent Line Equation

I need some help with these problems:

Find an equation of the tangent line to the curve at the given point:

1. y = (x-1)/(x-2), (3,2)

2.y = 2x^3-5x, (-1,3)

2. Originally Posted by dm10
I need some help with these problems:

Find an equation of the tangent line to the curve at the given point:

1. y = (x-1)/(x-2), (3,2)

2.y = 2x^3-5x, (-1,3)
do you know how to find the equation of a line if you are given the slope and a point the line passes through?

note that the derivative gives you the formula for the slope at any value of x

3. Originally Posted by Jhevon
do you know how to find the equation of a line if you are given the slope and a point the line passes through?

note that the derivative gives you the formula for the slope at any value of x
I don't know what to do.

4. Originally Posted by dm10
I don't know what to do.
for the first, apply the product rule: $\frac d{dx} \frac {f(x)}{g(x)} = \frac {f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

once you find the derivative, plug in x = 3. that will give you the slope of the line, that is, m, in y = mx + b. now you know m and you know (x,y) = (3,2) you can find b and hence the equation of the line.

do a similar thing for the second problem, except when finding the derivative, use the power rule which i told you about in your other post

try it

5. Originally Posted by Jhevon
for the first, apply the product rule: $\frac d{dx} \frac {f(x)}{g(x)} = \frac {f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

once you find the derivative, plug in x = 3. that will give you the slope of the line, that is, m, in y = mx + b. now you know m and you know (x,y) = (3,2) you can find b and hence the equation of the line.

do a similar thing for the second problem, except when finding the derivative, use the power rule which i told you about in your other post

try it
I do that but I keep getting the wrong answer.

6. Originally Posted by dm10
I do that but I keep getting the wrong answer.
show your work, i can tell you where you are messing up

7. $y-y_1=m(x-x_1)$ Your slope is the answer to the x value plugged in your derivative. The x value is the value plugged into your original equation, the y is the value of that.