# Finding the equation of the tangent line for the curve $f(x)=x^2+\frac{2}{x}$ parallel to $3y-2x=5$

Find the equation of the tangent line for the curve defined by

$$f(x)=x^2+\frac{2}{x}$$

That is parallel to

$$3y-2x=5$$

Since

$$y=\frac{5+2x}{3}$$

The slope of the parallel line is $\frac{2}{3}$, so the tangent line is also $\frac{2}{3}$.

The derivative of the curve function is

$$f'(x) = 2x - \frac{2}{x^2}$$

So if I solve

$$2x - \frac{2}{x^2} = \frac{2}{3}$$

I would get the points where there is a tangent line such that the slope is $\frac{2}{3}$, right?

But as far as I'm concerned, it's not the points what I want. It's the equation for the tangent line itself.

What should I do?

• Once you know the points where the slope matches you only need to find the point the curve (graph) passes through and write down the equation for the line through that point with the given slope. Commented Nov 8, 2015 at 8:10
• Getting $x_0$ from your equation, you substitute in $f$ to get $y_0$, then the equation of the tangent is $y-y_0= \frac{2}{3}(x-x_0)$ Commented Nov 8, 2015 at 8:13
• There are also several related questions in the list to the right.
– A.Γ.
Commented Nov 8, 2015 at 8:15

the equation of tangent line $$y=\frac{2}{3}x+b$$ if $(x_0,y_0)$ is a tangent point $$y_0=\frac{2}{3}x_0+b$$ $$b=y_0-\frac{2}{3}x_0$$
Your tangent line will have the same slope as $3y-2x = 5$, so it will have a form
$3y - 2x = b$
Once you find an $(x, y) = (x, f(x))$ of the points where the tangent line is 2/3, apply it to the equation above and solve for $b$, that will be your solution(s).