Review question

# When is this parabola's turning point nearest to the origin? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9506

## Solution

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

1. $a=0$,

2. $a=\pm1$,

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

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

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.

2. 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.

3. 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).

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