The diagram shows the curve y = x^2 + px + q. The points A(0, −5) and B(2, −9) lie on the graph. B(2, −9) is the minimum point.

1. Use point A to find $q$.

Since $A(0,-5)$ lies on the curve,
$-5=0^2+p(0)+q$
$q=-5$

2. Use point B to form an equation involving $p$.

Since $B(2,-9)$ lies on the curve,
$-9=2^2+p(2)+q$
$-9=4+2p+q$

3. Substitute $q=-5$ and solve for $p$.

Substitute $q=-5$:
$-9=4+2p-5$
$-9=2p-1$
$2p=-8$
$p=-4$

1. Find the gradient of line $BC$.

For $B(2,-9)$ and $C(7,16)$,
$m_{BC}=\frac{16-(-9)}{7-2}=\frac{25}{5}=5.$

2. Use the same gradient and point $A$ to find the required equation.

The required line is parallel to $BC$, so its gradient is also $5$.

It passes through $A(0,-5)$, so the $y$-intercept is $-5$.

Hence,
$y=5x-5.$