This was on my year 10 maths test and I gave up with 40 mins to complete:
Basically you were given the coordinates: y intercept : (0,10) 1 x intercept: (10,0) and y value of the vertex: +15
Can someone help me form a quadratic equation from just those coordinates?
THANKS
We can write it as
$$y=a(x-b)^2+c$$
where $c$ is the $y$-value at the vertex, $c=15$. Now we can make it pass through $(0,10) $ and $(10,0)$ and this will determine $a$ and $b$.
We get two equations:
$$10=a(0-b)^2+15$$
and $$0=a(10-b)^2+15$$
We need to solve the system
$$\begin{align}ab^2+15&=10\\a(b^2-20b+100)+15&=0\end{align}$$
So $-20b+100=0$, i.e. $b=5$. Hence $a=15/25$.