Finding a distance between a variable point and a fixed point on a parabola

Question

Question

The parabola $y=x^2-x-2$ intersects the $x$-axis at points $A$ and $B$, the $y$-axis at point $C$, and its vertex is at point $M$. If there is a variable point $P$ on the parabola to the right of $M$, prove that $AP>AC$.

My Working

$A$ and $C$ are fixed points, and so I can find $AC$, which is $\sqrt{2^2+1^2}=\sqrt5$. I'm not sure about what to do next, but there might be something about the parabola strictly increasing past $M$?

Edit: After getting some tips, I constructed a circle with centre $A$ and radius $AC$, which has the equation $(x+1)^2+y^2=5$. To find the points of intersection of the circle and the parabola, I needed to solve the set of questions:

\begin{align} y&=x^2-x-2\\ (x+1)^2+y^2&=5 \end{align}

Squaring both sides of the first equation and substituting the result and simplifying gives:

$$x^4-2x^3-2x^2+6x=0$$

We can factorise $x$ out, and this gives

$$x\left(x^3-2x^2-2x+6\right)=0$$

However, the cubic in the brackets cannot be factorised, and I'm not sure how to proceed.

Desmos link: https://www.desmos.com/calculator/kbmpxsdz9n

Answer 1

If you are allowed to use derivatives, it becomes relatively easy to find the local minimum and maximum of $x^3-2x^2-2x+6$, show that they are both positive, and use this fact to show that $x^3-2x^2-2x+6$ is positive for all positive values of $x$.

But if this is really a pre-calculus problem, you will need other tools.

The fact that they only ask to proof $AP > AC$ when $P$ is to the right of $M$, rather than when $P$ is to the right of the $y$-axis, suggests that you use the point $M$ in some way. For example, construct a line $\ell$ through $M$ and $B$. Prove that that for every point $Q$ on this line, $AQ > AC$. Then show that for any point $P$ on the parabola between $M$ and $B$, the line segment $AP$ passes through the line $\ell$, and if you let $Q$ be the point of intersection then $AP > AQ > AC$.

For $P = B$ and points to the right of $B$, the fact that the difference in $x$ coordinates is already greater than $\sqrt5$ is sufficient.

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You could convert this to a proof for all $P$ to the right of the $y$-axis by using the line through $C$ perpendicular to $AC$ rather than the line through $M$ and $B$.

Answer 2

Sometimes representing a polynomial differently brings clarity, like so: $$x^3-2x^2-2x+6 = x(x-2)^2+2(x-\tfrac32)^2+\tfrac32$$

Hence the cubic you are concerned about cannot have positive roots...