Equation of the tangent line?? I was wondering if any of you knew how to find the equation of the line tangent to the curve $y = x^3-5x+2$ at point $(2,0)$ in standard form.
I personally got $7x - y - 14 = 0$. However, that is apparently incorrect.
Does anyone know what I may be doing wrong? Thank you in advance!
 A: This is correct. To check it, you need first to find the slope of the tangent line: from $y=x^3-5x+2$, we have $y'=3x^2-5$. For $x=2$, this gives $y'=12-5=7$.
Therefore, the slope-point equation of the tangent line is $y=0+7(x-2)$ which is equivalent to $y=7x-14 \iff 7x-y-14=0$.
A: Your answer is correct, but can be written more explicitly as $y=7x-14$. If $f(x)=x^3-5x+2$, then $f'(x) = 3x^2-5$. In general, the equation of the tangent to the point $(a,f(a))$ is given by
$$
y-f(a) = f'(a)(x-a) \, .
$$
Here, $a=2$, $f(a)=0$, and $f'(a)=f'(2)=7$. Hence, the equation of the tangent line is
$$
y=7(x-2) \implies y=7x-14 \, .
$$
A: Your answer is correct. However, in the hope that this will be helpful to you later on, I am going to provide you with a standard algorithm that helps you determine the tangent plane at a certain point on a level set of $F : \mathbb{R}^n \rightarrow \mathbb{R}.$ Assume that $F$ is "nice enough" (for example, take it to be $C^1$). Consider the level set $F(x_1, \dots, x_n) = c,$ where $c \in \mathbb{R}.$ Then it is well known fact that the normal at a point $y \in F^{-1}(c)$ is given by the gradient of $F$ at that point, $\nabla F (y)$ (for a proof, consider a curve on the level set passing through your point, $\gamma$; then you have that $F(\gamma(t)) = c$ and just use the chain rule to obtain this lemmata). It follows that your tangent plane at $y$ is given by $\nabla F(y) \cdot x = 0$ (or a plane parallel to it, rather).
In this particular case, we have $F(x, y) = y - x^3 + 5x - 2,$ so that $\nabla F(x, y) = (-3 x^2 + 5, 1).$ In particular, $\nabla F(2, 0) = (-7, 1).$ Thus the equation of the tangent at $(2, 0)$ will be $-7x + y + c = 0.$ Since we want $(2, 0)$ to be in this plane, we get $c = 14.$ Hence the answer: $-7 x + y + 14 = 0.$ I hope this helps. :)
