Prove that all the tangents of a function do not touch a given point

Question

A quick question that I'm stuck on:

Prove that the point $(2,1)$ does not touch any of the tangent lines of $f(x)= x^2 - 4x$.

Answer 1

The tangent line to $f(x)$ at $x=c$ is given by $$y_c=f'(c)(x-c)+f(c)=(2c-4)x-c^2.$$

Suppose, for some $c$, this line contains $(2,1)$. Then $1=(2c-4)2-c^2$. Can you take it from here?

Answer 2

More generally, we want to show that the tangent to the curve is below the curve. Since your point is above $(2, f(2)) =(2, 0)$, all the tangents will be below it.

The tangent at $(a, f(a))$ is $y = f(a)+f'(a)(x-a) $.

Putting in $x=b, y=f(b)$, we want to show that when $b \ne a$ we have $f(b) \gt f(a)+f'(a)(b-a) $ or $f(b)- f(a) \gt f'(a)(b-a) $.

If $f(x) =x^2-4x $, $f'(x) = 2x-4 $ so

$\begin{array}\\ f(b)-f(a) &=b^2-4b-(a^2-4a)\\ &=b^2-a^2-4(b-a)\\ &=(b-a)(b+a-4)\\ \end{array} $

and $(b-a)f'(a) =(2a-4)(b-a) $ so

$\begin{array}\\ f(b)-f(a)-(b-a)f'(a) &=(b-a)(b+a-4)-(b-a)(2a-4)\\ &=(b-a)(b+a-2a)\\ &=(b-a)^2\\ &\gt 0\\ \end{array} $

Note that this holds for any $f(x) =x^2+ux+v$.

You can also show that $f''(x) > 0$ implies that $f(b)-f(a) \gt (b-a)f'(a) $.

Answer 3

Any line through the point $(2,1)$ has an equation of the form $ax+by=2a+b$ for $a$ and $b$ not both $0$. If this is tangent to the parabola, then the coefficients $a$ and $b$ satisfy the dual conic equation $$\begin{bmatrix}a&b&-2a-b\end{bmatrix} \begin{bmatrix}1&0&-2\\0&0&-\frac12\\-2&-\frac12&0\end{bmatrix}^{-1} \begin{bmatrix}a\\b\\-2a-b\end{bmatrix} = a^2+20b^2 = 0.$$ The only real solution is $a=b=0$, therefore no line through $(2,1)$ is tangent to the parabola.