202210191554 Solution to 1969-CE-AMATH-II-XX

Find the $x$ -coordinate of the point on the curve $y=ax^2+bx+c$ where the tangent is horizontal.

Roughwork.

The gradient $y'(x)$ of the curve $y(x)$ is

$y'(x)=2ax+b$ .

When the tangent is horizontal,

$\begin{aligned} y'(x) & =0 \\ 2ax+b & =0 \\ x & = -\frac{b}{2a} \\ y & = a\bigg( \frac{-b}{2a}\bigg)^2+b\bigg( \frac{-b}{2a}\bigg)+c \\ & = \dots \\ \end{aligned}$

物理後生﹑掣數學之肘.
小子算數﹑指物理之月.

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31