A quadratic function f satisfies f(x) <= 0 exactly on the interval -1 <= x <= 5, and its minimum value is -18. Solve the inequality f(x) <= 6.

Step-by-step solution

1. 1

   Read the roots and the direction of opening straight off the solution set. $f(x)\le0$ holds exactly on a closed bounded interval, so $f$ vanishes at the endpoints and is negative strictly between them. That means $-1$ and $5$ are the two roots and the parabola opens upward (if it opened downward, $f\le0$ would be the two outer rays instead). Therefore

   $f(x)=a(x+1)(x-5),\qquad a>0.$

2. 2

   Use symmetry to locate the vertex without completing the square. For an upward parabola the minimum sits midway between the roots:

   $x_{\text{vertex}}=\frac{-1+5}{2}=2.$

   This is the only place the stated minimum value can occur.

3. 3

   Turn the minimum value into an equation for $a$.

   $f(2)=a(2+1)(2-5)=-9a$

   $-9a=-18\quad\Longrightarrow\quad a=2$

   The sign check passes: $a=2>0$, consistent with the upward opening deduced in step 1. So

   $f(x)=2(x+1)(x-5)=2x^2-8x-10.$

4. 4

   Solve $f(x)\le 6$. Move everything to one side and divide by the positive leading coefficient (dividing by a positive number does not flip the inequality):

   $2x^2-8x-10\le6\;\Longrightarrow\;2x^2-8x-16\le0\;\Longrightarrow\;x^2-4x-8\le0.$

5. 5

   Find the roots of the boundary equation and pick the correct side.

   $x=\frac{4\pm\sqrt{16+32}}{2}=\frac{4\pm4\sqrt3}{2}=2\pm2\sqrt3$

   Since $x^2-4x-8$ opens upward, it is $\le0$ between its roots:

   $2-2\sqrt3\le x\le 2+2\sqrt3\qquad(\text{approximately }-1.464\le x\le5.464).$

6. 6

   Sanity-check the answer against the given data. The new interval is centred on $x=2$, the same axis of symmetry, and it strictly contains $[-1,5]$ — it must, because the level $6$ is above the level $0$ on an upward parabola. Substituting the endpoint: $f\left(2+2\sqrt3\right)=2\left(3+2\sqrt3\right)\left(-3+2\sqrt3\right)=2\left(12-9\right)=6$, exactly the boundary value.