Is there a problem in using this method to find a tangent line to a function? I've been asked to find the equation of the tangent line to the point $(1,f(1))$ for the function $f(x)=x^3$. I did the following:


*

*As $f'(x)=3x^2$, then the slope of $f$ at the given point is $3$.

*Then the equation of the line is $y=3x+b$; now I need to find $b$.

*I found I can substitute $y$ for $f(1)$ and $x$ for $1$, this yields: $b=-2$.

*Then the equation of the line tangent to $(1,f(1))$ is $y=3x-2$
It worked via inspection for all examples I did. But I remember the professor taught us another way which seemed more complicated than what I'm doing. I'm not sure if both ways are equivalent; I'd like to know if there is a problem in doing this procedure the way I'm doing.
 A: The "other way" you mention is probably the point-slope form of the line,
$$y=m(x-x_1)+y_1$$
or sometimes written
$$y=y_1+m(x-x_1)$$
In that method, the slope $m$ is found in the same way you found it. Then you substitute the $m$ and the coordinates $x_1$ and $y_1$ of the given point. Then you simplify the expression on the right side of the equation.
Your method also works just fine for this kind of problem. Its advantage is that you had fewer equations to memorize: the slope-intercept form of the line is known by practically everyone and you did not need the point-slope form at all.
However, the point-slope form also has advantages. For example, it is one step to finding the Taylor series of the function at the point, whereas your method is useless for that. The point-slope also helps in understanding differentials and error approximation.
In summary, your method works fine for this particular kind of problem, but you really should also work to understand the professor's method so you can do later, more difficult problems.
A: Your method is legitimate because the tangent line passes through the point $(1, f(1))$.
There is another way to do it. Let $g: (x,y) \mapsto y-x^{3}: \mathbb{R}^{2} \to \mathbb{R}$. Then the zero set $g^{(-1)}\{ 0 \}$ is the graph of $f$. But $\nabla g(x,y) = (-3x^{2},1)$ for all $x,y \in \mathbb{R}$, so $\nabla g(1, 1) = (-3,1)$, which is a vector orthogonal to the tangent line through $(1,1)$. So the tangent line is simply the set of all $(x,y) \in \mathbb{R}^{2}$ such that $\nabla g(1,1)\cdot (x-1, y-1) = y - 3x + 2 = 0$.
A: Your procedure is correct. 
If we know the point of tangency then we can use point-slope form of equation in the following steps (somewhat different from OP's procedure)  
Given $$f(x)=x^3\implies f'(x)=3x^2$$
$$\implies f(1)=(1)^3=1 $$


*

*The point of tangency is $(1, f(1))\equiv (1, 1)$

*Slope of the tangent at the point $(1, 1)$ is 
$$m=[f'(x)]_{(1, 1)}=3(1)^2=3$$

*Hence, the equation of the tangent having slope $m=3$ & passing through $(1, 1)$ is given by point-slope form $$\color{blue}{y-y_1=m(x-x_1)}$$
Setting the corresponding values $$y-1=3(x-1)$$ $$\color{red}{y=3x-2}$$

A: Perhaps your professor used the formula
$$m=\frac{\Delta y}{\Delta x}$$
First, you found $m=3$, thus $3=\Delta y / \Delta x$ or alternatively
$$3 = \frac{y-y_1}{x-x_1}=\frac{y-1^3}{x-1}$$
Which can be restated in point-slope form
$$y-1=3(x-1)$$
and restated again in slope-intercept form
$$y=3x-2$$
And to recap my preceding discussion
$$m=\frac{\Delta y}{\Delta x}$$
Before I finish, though, lets just look at what you did symbolically.
 find the equation of the tangent line to the point (a,f(a)) in the function f.
Step 1 is just the statement that
$$m=f'(a)$$
continuing ahead to step 2, we have
$$y=mx+b=f'(a)x+b$$
In step 3, we substituted $a$ for $x$ and $f(a)$ for $y$
$$f(a) = f'(a)a + b$$
$$b = f(a) - af'(a)$$
Then we put it all together in step 4 to get the tangent line (in slope-intercept form)
$$y = mx+b=f'(a) x + \bigg(f(a) - af'(a)\bigg)$$
in point-slope form
$$y=f'(a)(x-a)+ f(a)$$
and in slope-definition form
$$f'(a)=\frac{y-f(a)}{x-a}$$
And just  to reiterate the only formula here that you ever need to memorize
$$m = \frac{\Delta y}{\Delta x}$$
