In the xy-plane a parabola has vertex (9, -14) and crosses the x-axis twice. If its equation is written as y = ax^2 + bx + c, which values could a + b + c take?

Step-by-step solution

1. 1

   Recognise what a + b + c means. Substituting $x=1$ into $y=ax^2+bx+c$ gives $a(1)+b(1)+c=a+b+c$. So

   $a+b+c=y(1),$

   the height of the parabola at $x=1$. You never need to find $b$ and $c$ individually — this identity is the whole shortcut.

2. 2

   Write the parabola in vertex form. With vertex $(9,-14)$,

   $y=a(x-9)^{2}-14.$

3. 3

   Evaluate at x = 1.

   $a+b+c=y(1)=a(1-9)^{2}-14=64a-14.$

4. 4

   Pin down the sign of a from the two x-intercepts. The vertex sits at $y=-14$, below the $x$-axis. A parabola whose lowest point is below the axis can only cross the axis twice if it opens upward, so

   $a>0.$

   (If $a<0$ the vertex would be the maximum, and the entire curve would stay at or below $y=-14$, never touching the axis.)

5. 5

   Convert the constraint on a into a constraint on the sum. Since $a>0$,

   $64a>0\quad\Longrightarrow\quad a+b+c=64a-14>-14.$

   So any number strictly greater than $-14$ is attainable, and no number $\le -14$ is.

6. 6

   Confirm with a concrete parabola. Take $a=1$: $y=(x-9)^2-14=x^2-18x+67$, so $a+b+c=1-18+67=50=64(1)-14$ ✓. It has vertex $(9,-14)$ and roots $9\pm\sqrt{14}$ — two $x$-intercepts, as required. Taking $a=\tfrac{1}{64}$ instead gives $a+b+c=-13$, showing values just above $-14$ are reachable too.