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A quadratic polynomial $p(x)$ is such that $p(x)$ never takes any negative values. Also, $p(0)=8$ and $p(8)=0$. What would $p(-4)$ be?

I tried doing it by taking the minimum value as zero that is the vertex of the polynomial at $8$. How do we go about after finding out values of $a$ and $b$ in the standard form of the equation $ax^2 + bx + c$ ? Could this be done in a shorter way graphically?

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  • $\begingroup$ If $p(x)$ never takes any negative values, wouldn't $p(-4)$ be undefined? Or invalid? $\endgroup$ – Brian J Jun 1 '15 at 19:02
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    $\begingroup$ @BrianJ I think he means $p(x) \geq 0 \forall x$. $\endgroup$ – Johanna Jun 1 '15 at 19:04
  • $\begingroup$ Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Martin Sleziak Jun 2 '15 at 5:04
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Here's a way to solve this without needing as much mathematical insight as Barry's answer. If the polynomial is $$p(x) = ax^2 + bx + c,$$ then from $p(0) = 8$ we get $$0a + 0b + c=8,$$ from $p(8) = 0$ we get $$64a + 8b + c = 0,$$ and from $p(x)$ never taking any negative values, it follows that $x=8$ is a minimum, and so $p′(8) = 0$, and since $p′(x) = 2ax + b$, then we get $$16a + b + 0c = 0.$$ This gives three simultaneous equations in three variables, which give the coefficients $$a={1\over 8}, b=-2, c=8.$$

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Since $p(x)$ is always non-negative, and $p(8) = 0$, it follows that that must be the vertex. The standard form for a parabola is $a(x-h)^2 + k$, and we already know that $(h,k)$ is $(8,0)$. That gives us $p(x) = a(x-8)^2$.

Now simply plug in $p(0) = 8$ to get $a = \frac{1}{8}$ to find $p$:

$$p(x) = \frac{1}{8}(x-8)^2$$

So:

$$p(-4) = 18$$

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