1. ## Minimum point on a quadratic curve

How do I find the minimum point on a quadratic curve other than using the completing the square method e.g. I'm presented with x^2+10x+21 and it cannot be done using the above method.

Please help me, otherwise how would I find a maximum point (by any method lower than A-Level).

Thansk.

2. Originally Posted by Mukilab
How do I find the minimum point on a quadratic curve other than using the completing the square method e.g. I'm presented with x^2+10x+21 and it cannot be done using the above method.

Please help me, otherwise how would I find a maximum point (by any method lower than A-Level).

Thansk.
You can complete the square on any quadratic - even the quadratic formula is derived by completing the square. I suspect they want you to complete the square because the only other way is to find $\displaystyle \dfrac{dy}{dx} = 0$ and then check the sign of $\displaystyle \dfrac{d^2y}{dx^2}$. In other words calculus which is not lower than A level.

Completed the square form is $\displaystyle (x+k)^2 + h$. The lowest point on the graph is equal to $\displaystyle h$ because $\displaystyle (x+k)^2 \geq 0$

$\displaystyle ax^2+bx+c$ into completed square form is $\displaystyle \left(x+\dfrac{b}{2a}\right)^2 - \dfrac{b^2}{4a^2} + c$.

Aligning that with the $\displaystyle (x+k)^2 + h$ form we get $\displaystyle k = \dfrac{b}{2a}$ and $\displaystyle h = \dfrac{b^2}{4a^2} + c$

In your case you have $\displaystyle a=1$ , $\displaystyle b = 10$ and $\displaystyle c=21$

My answer is in the spoiler but if you have any issues getting there please post your workings

Spoiler:
$\displaystyle (x+5)^2-4$ - this means -4 is the minimum point and this happens when $\displaystyle x = -5$ so the minimum point is $\displaystyle (-5,-4)$

3. My thanks, that formula of k= and h= is very clear and concise and will serve me throughout the future year I suspect

I guess I still have to learn the A-Level method soon

Thank you!