# Thread: find each point at which the tangent line to the curve...

1. ## find each point at which the tangent line to the curve...

What do I do? Take y' of the first equation and then set it equal to the slope of the second equation and solve for x? answer is A.

This is what I got:
$
y1'=2-4x^{-2}$

$y2=-2x+6$

$2-4x^{-2}=-2x+6$
$2(2x^{-2}-x+2)$

Did I take the wrong route/ what would I do next?

2. Originally Posted by hazecraze

What do I do? Take y' of the first equation and then set it equal to the slope of the second equation and solve for x? answer is A.

This is what I got:
$
y1'=2-4x^{-2}$

$y2=-2x+6$

$2-4x^{-2}=-2x+6$
$2(2x^{-2}-x+2)$

Did I take the wrong route/ what would I do next?
You got the correct y1'. You're making the rest too complicated. If the tangent line of y1 is parallel to y2, they simply have the same slope. If you have the form y=mx+b, m is the slope. So the slope of y2 is -2. Set y1' = -2 and solve for the x values.

3. check your derivative of the first one.
y' should be 2-4x^-5.

When one line is parallel to another that means they have the same slope. using slope intercept form

(y_0-y1)=m(x-x1) you can plug the points into the equation.

4. Originally Posted by pham07
check your derivative of the first one.
y' should be 2-4x^-5.

When one line is parallel to another that means they have the same slope. using slope intercept form

(y_0-y1)=m(x-x1) you can plug the points into the equation.
I think you made an error with y'.

$\frac{d}{dx} 2x+\frac{4}{x}= \frac{d}{dx} 2x+4x^{-1}=2-\frac{4}{x^2}$