Answer
$y=3x - 2$
$y=3x + 2$
Work Step by Step
You begin by finding the derivative of the given function: $x^{3}$.
You can begin by using the limiting process: $\lim\limits_{\Delta x \to 0} \frac {f(\Delta x+x)-f(x)} {\Delta x}$
Substituting the given function in: $\lim\limits_{\Delta x \to 0} \frac {(x+\Delta x)^{3}-x^{3}} {\Delta x}$
This can now be multiplied out: $\lim\limits_{\Delta x \to 0} \frac {x^{3}+3x^{2}\Delta x + 3x \Delta x^{2} + \Delta x^{3} - x^{3}} {\Delta x}$
Delta x can then be divided out and the x cubes will cancel: $\lim\limits_{\Delta x \to 0} 3x^{2} + 3x \Delta x + \Delta x^{2}$
Applying the limit gives you $3x^{2}$
This must then equal the slope of the given equation because the lines are parallel, putting the equation in slope intercept form yields: $y = 3x +1$ therefore the slope is $3$ and $3$ must equal the derivative of the function.
Dividing both sides by $3$ and taking the square root yields Positive and negative $1$.
$f(1)=1^{3}$ and $f(-1)=(-1)^{3}$, therefore, the tangent line passes through ($1, 1)$ and $(-1, -1)$. Now the equation is left. $y=3x+b$
$1 = 3(1) +b $
$b=-2$
$-1 = 3(-1) +b $
$b=2$
And the solution can now be written as $y=3x -2$; $y=3x+2$
