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 have this equation where the question says,

What is the smallest value, to 2 decimal places, in the image set of the function


a = 3  b =12 c =180

so i used this formula to get this answer 189.67,


then i simply substitute the values is that correct?

share|cite|improve this question
How and why is this [elementary-set-theory]?? – Asaf Karagila Sep 4 '12 at 8:06
up vote 2 down vote accepted

$f(x)=ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}≥c-\frac{b^2}{4a}$ if $a>0$ as $(x+\frac{b}{2a})^2≥0$ for real $x,a,b$.

So, the minimum value of $f(x)=ax^2+bx+c$ is $c-\frac{b^2}{4a}$ if $a>0$

Putting $a=3,b=12,c=180$, $c-\frac{b^2}{4a}=180-\frac{(12)^2}{4\cdot 3}$ $=180-12=168=168.00$ (with 2 decimal places precision)

The extreme value of $ax^2+bx+c$ can be calculated using another approach apart from differentiation as follows:

Let $y=ax^2+bx+c\implies ax^2+bx+c-y=0$

As $x$ is real, $(-b)^2≥4\cdot a\cdot(c-y)\implies \frac{b^2}{4a}≥c-y\implies y≥c-\frac{b^2}{4a} $

So, $ax^2+bx+c=y≥c-\frac{b^2}{4a}$

share|cite|improve this answer

You did not substitute into the formula correctly.

$\frac{4(3)(180) - 12^2}{4(3)} = \frac{2160 - 144}{12} = \frac{2016}{12} = 168$

which indeed is the correct answer. Below is a hint on an alternative way to solve this problem.

$\textbf{Hint}$ : Find the minimum of the function $3x^2 + 12 x + 180$ by differentiating.

Move cursor over box for more details.

Given any differentiable function $f : \mathbb{R} \rightarrow \mathbb{R}$. The values $x$ such that $f(x) = 0$ are local maximums or minimums of the function. The derivative of this function is $6x + 12$. The only zero of this function is $-2$. Now it very to check that the function obtains a minimum at $-2$. Substituting $-2$ into the function, you see that $3(-2)^2 + 12(-2) + 180 = 168$ is the minimum value.

Actually using the method above, you can derive the formula. Again I put into box in case you want to figure it out yourself

Let $ax^2 + bx + c$ denote an arbitrary degree two polynomial. Using the same idea as above, the minimum or maximum occurs at the zeros of $2ax + b$. That is, it occurs at $\frac{-b}{2a}$. Substituting back into the original function, you get that the minimum or maximum is
$$a\left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c$$
$$= \frac{ab^2}{4a^2} + \frac{-b^2}{2a} + c$$
$$= \frac{b^2}{4a} + \frac{-2b^{2}}{4a} + \frac{4ac}{4a}$$
$$= \frac{4ac - b^2}{4a}$$
This is your formula.

share|cite|improve this answer
hmm interesting but the hint given is 2 decimal places so does that mean anything? – JackyBoi Sep 4 '12 at 8:05
@JackyBoi Giving the answer to two decimal place is not really a hint. It is just direction on how to specify the answer. In any case, 168 is the answer up to two decimal places. – William Sep 4 '12 at 8:27

Yes, your formula is correct. The number isn't though, it should be $168$, or $168.00$ to 2 decimal places.

There is the so-called vertex form for any parabola given by $ax^2+bx+c$, which is


You get there by Completing the square.

In this form, one can see directly that the minimal point is


so the $y$-coordinate gives you the smallest value the parabola does attain.

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.