Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to determine the maximum value for y = ax^2 + bx + c, where I know the coefficients and the upper and lower x values.

Say the input values are:

  • a = 5
  • b = 1
  • c = 2
  • x lower limit = -5
  • x upper limit = 5

Given these input, how do I determine the the maximum value for the quadratic equation above?

My goal is to implement a function in a computer programming language that has a signature such as funcMax(int a, int b, int c, int xUpper, int xLower). The part that is confusing me is that I have two x values for this signature but one x in the formula above.

But computer programming aside for now, mathematically, how do I get the max value for the above equation, using the above input (with two x values)?

share|cite|improve this question
For this quadratic function, the global maximum happens either at the two ends or at a local maximum. – peterwhy Nov 18 '13 at 2:58
up vote 1 down vote accepted

For this quadratic function, the global maximum happens either at the two ends or at a local maximum.

The first two cases are simple, you have to only compare the $y$ evaluated at the upper and lower limits of $x$ respectively.

And there is a local maximum only if the $a$ value is negative, shown by the second derivative test. When this is the case, the maximum value is given by taking $x = -\frac b{2a}$. (You can verify this by completing square or first derivative test.) Then you have to check if this $x$ is within your domain given by the upper and lower limits.

If I write your required function out:

$$\begin{align*}&\text{funcMax}(a,b,c,x_{Upper}, x_{Lower}) \\ =& \begin{cases} \max(ax_{Upper}^2 + bx_{Upper}+c, ax_{Lower}^2 + bx_{Lower}+c,\frac{b^2}{4a}-\frac{b^2}{2a}+c) & \text{if }a\ne0\text{ and }x_{Upper}\le-\frac{b}{2a}\le x_{Lower}\\ \max(ax_{Upper}^2 + bx_{Upper}+c, ax_{Lower}^2 + bx_{Lower}+c) & \text{otherwise} \end{cases}\end{align*}$$

share|cite|improve this answer

In effect, you would like to determine the maximum value of the function $f \colon [-5,5] \to \mathbf{R}$ defined as $$ f(x):= 5x^2+x+2$$ for all $x$ in $[-5,5]$. We can do this using the methods of calculus as follows:

The function $f$ is continuous on $[-5,5]$ and differentiable on $(-5,5)$. Differentiate $f$, we find that $$f^\prime(x) = 10x+1. $$ Thus $f^\prime = 0$ for $x=-1/10$ only; whereas $f^\prime < 0$ for $x < -1/10$ so that $f$ is a decreasing function on the interval $(-\infty,-1/10]$; and $f^\prime > 0 $ for $x > -1/10$ so that $f$ is an increasing function on the interval $[-1/10, +\infty)$.

Thus it follows that $f(-5) > x > f(-1/10)$ for all $x$ such that $-5< x < -1/10$; in other words, $f(-5) = 5 \cdot (-5)^2 + (-5) + 2 = 122 \geq f(x) $ for all $x$ in the interval $[-5,-1/10]$.

Similarly, $f(5) = 132 \geq f(x) $ for all $x$ in the interval $[-1/10,5]$.

Hence the maximum value of the given quadratic is $132$ when $x=5$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.