Exercise 367

The graph of $y=x^2+2x-24$ intersects the $x$-axis at $A,C$ and the $y$-axis at $B$. What is $\tan(\angle ABC)$?

Solution

$x^2+2x-24=0\Rightarrow x=4$ or $x=-6$, so $A=(-6,0),C=(4,0),B=(0,-24)$.
$\vec{BA}=(-6,24),\vec{BC}=(4,24)$: $\tan(\angle ABC)=\dfrac{|(-6)(24)-(24)(4)|}{(-6)(4)+(24)(24)}=\dfrac{240}{552}$.
Simplifying: $\boxed{\dfrac{10}{23}}$.