Equation of a tangent to the graph of a function parallel to a line

Question

Please help me find the answer to this question. Thanks.

What is the equation of a tangent to the graph of a function $y=x-\frac{1}{x^2}$ which is parallel to the line $y=3x$?

Update: found the slope of the line, $3$.

According to one of the answers $\frac{dy}{dx}=3$, but how? If the slope $=3$, shouldn't the derivative be $0$.

I don't understand the "parallel to" part. I know how to find the slope of the tangent, but do not know what is meant by "the equation of a tangent to the graph".

Answer 1

HINT:

$$\frac{dy}{dx}=1+\frac2{x^3}$$ which needs to be $=3$

This will give us the value of $x, x_1$(say) and consequently the value of $y, y_1$

Now the equation of the required tangent will be $$\frac{y-y_1}{x-x_1}=3$$

Answer 2

The derivative of the function $f(x)=x-\frac{1}{x^2}$ is $f'(x)=1+\frac{2}{x^3}$. The tangent line to the graph of $f(x)$ at the point $x=a$ has slope $f'(a)$, so it will be parallel with the line $y=3x$ if and only if $f'(a)=3$. So, you are looking for the real numbers $a$ for which $$1+\frac{2}{a^3}=3$$

Answer 3

The slope of the line is $3$ . So, $\frac{dy}{dx}=3$

$1+\frac{2}{x^3}=3$$\implies x=1$ and $y=0$

Equation of the line is $(y-0)=3(x-1)$ $\implies y=3x-3$