Find an equation for the tangent line to $f(x)=3x^2-\pi^3$ at $x=4$ 
Find an equation for the tangent line to $f(x)=3x^2-\pi^3$ at $x=4$

Here is my work so far
$f'(x) = 6x$
$f'(4) = 24$
But if I treat $\pi$ like a variable, 
$f'(x) = 6x - 3\pi^3 $
$f'(4) = 24-3\pi^3 $
The answer to this problem is $$ y=24x−48−\pi^3 $$
but I am unsure as to how I should get to this answer. Any hints would be appreciated. 
 A: Remember that an equation of a line with slope $m$ passing by $(x_1,y_1)$ is $y-y_1=m(x-x_1)$, here we have
$$x_1=4,\quad y_1=f(x_1)=f(4)=3(4)^2-\pi^3=48-\pi^3\quad\text{ and }\quad m=f'(4)=24$$
It follows that an equation for the tangent line is
$$y-(48-\pi^3)=24(x-4)$$
Which es equivalent to
$$\boxed{\color{blue}{y=24x-48-\pi^3}}$$
A: As you had correctly determined, the slope of the line is $24$.  
Then, simply write
$$y=24x+b$$
where $b$ is the $y$ intercept. To find the intercept we make use of the fact that the tangent line actually touches the curve at $x=4$, $y=3\times 4^2-\pi^2=48-\pi^3$.  Therefore, 
$$b=(48-\pi^2)-24(4)=-48-\pi^3$$.
Finally, we have that the equation of the tangent line is
$$\bbox[5px,border:2px solid #C0A000]{y=24x-48-\pi^3}$$
A: We have $$f(x)=3x^2-\pi^3\implies f'(x)=6x$$
Hence, $$f'(4)=6(4)=24$$ Now, setting $x=4$ in the given equation $f(x)=3x^2-\pi^3$, we get $$f(4)=3(4)^2-\pi^3=48-\pi^3$$
Hence, the equation of the line is given by point-slope form $$y-(48-\pi^3)=24(x-4)$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{y=24x-48-\pi^3}}$$
