When is this parabola's turning point nearest to the origin?

When is this parabola's turning point nearest to the origin?

Question

The turning point of the parabola $y=x^2-2ax+1$ is closest to the origin when

1. $a=0$,

---

1. $a=\pm1$,

---

1. $a=\pm \frac{1}{\sqrt{2}}$ or $a=0$,

---

1. $a=\pm \frac{1}{\sqrt{2}}$.

Suggestion

First we need to find the turning point of the parabola. How might we do this?

How can we work out the distance between this point and the origin?

Solution

One approach is to think about each case in turn and rely on transformations. For ease we will let $f(x)=x^2$

1. If $a=0$ we have $y=f(x)+1$: the turning point is $1$ unit from the origin.

---

1. If $a=\pm1$ I can recognise this as an expanded square bracket of $x\pm1$ so I have $y=f(x\mp1)$: again the turning point is $1$ unit from the origin.

---

1. can’t be correct from a logical point of view: since in cases (a) and (b) the turning points are the same distance from the origin then (c) must also include $a=\pm1$ if $\pm \frac{1}{\sqrt{2}}$ and $a=0$ are both options (i.e. the turning point has the same proximity in both cases), so $\pm \frac{1}{\sqrt{2}}$ is different and therefore smaller (because if it were larger there would need to be the answer $\, \text{"} a=0$ or $a=\pm1 \text{"}\,$ available).

Hence the answer is (d).

A second approach is to find the turning point of the parabola. Completing the square, we have

$$\begin{align*} y &= x^2 - 2ax + 1 \\ &= (x - a)^2 + 1 - a^2, \end{align*}$$

so the minimum occurs when $x = a$ and then $y = 1 - a^2$. So the turning point is at $(a, 1 - a^2).$

Now the distance from the turning point to the origin is given by

$$\begin{align*} D &= \sqrt{a^2 + (1-a^2)^2} \\ &= \sqrt{a^4 - a^2 + 1} \\ &= \sqrt{\left(a^2 - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}}, \end{align*}$$

where we have again completed the square.

So this distance is minimised when $\left(a^2 - \frac{1}{2}\right)^2=0,$ which is when $a=\pm\dfrac{1}{\sqrt{2}}$.

So the answer is (d).

Being able to minimise or maximse something is a useful tool (think costs, profits, time etc.). We could minimise our distance here because we were able to complete the square, but in order to minimise different functions we will need to explore ideas that appear at the Calculus of Powers station and beyond.

This GeoGebra file helps us to see what is happening.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/YGjuzvmT/width/400/height/600/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file