Other Points Where Tangent Line Intersects Graph

Question

Q: For each a ∈ $R$ find any other points at which the tangent line ($y = 3a^2 -48$) intersects the original graph ($x^3 - 48x + 2$). Hint: $f(x) − f(a)$ is divisible by $x − a$

Does this just mean whenever $x-a \not= 0$?

Answer 1

First of all, note that you are given a $y=k$ type line where $k$ is a constant.

For this line, the slope comes out to be $0$. Time to find out at which point the slope is $0$:

$$ \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3}-48x+2 \right ) = 3x^{2}-48 $$

Given that it is equal to $0$, there are two values of $x$, that is $x=\pm 4$. The two points are $ (4, -126)$ and $(-4,130)$, so the two lines are now $y=130$ and $y=-126$.

In order to find the where the graph cuts again we just plug in these value in $f(x)=126 \text{ or } 130$. These give the value $x=8$ for $130$ and $x=-8$ for $-126$.

Now note that $y=3a^2-48=-126$ cannot be used because of the fact that $a$ is not real here. So the final answer is $(8, 130)$.

As from the hint, I suggest you to read Remainder Factor Theorem. And in order to solve assume a new function $g(x)=f(x)-f(a)$, and apply remainder theorem.