Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the minimum value of $x^2+12x$?

I do not know what is meant by the minimum value.

share|improve this question
Why did you write that your question requires moderator attention? –  Alex Becker Jan 2 at 4:57
"Minimum value" refers to the smallest value that you can get from $x^2 + 12x$. –  T. Bongers Jan 2 at 4:57
How is defined the minimum of a function ? What did you try ? –  Claude Leibovici Jan 2 at 4:58
I've added a more appropriate tag. For future information, to get a moderator's attention you should flag the question, using the "flag" button at the bottom of the question. But you don't need a moderator to edit tags; most users can do so. So if you're not sure how to tag a question just add a comment to that effect. –  Alex Becker Jan 2 at 5:03
This edit pileup was ridiculous. –  Alex Becker Jan 2 at 5:06

6 Answers 6

up vote 2 down vote accepted

I realize this has been answered but here is my two bits worth.

Look at $x^2+12 x$. It is an expression. Now pick a value for $x$, say $x=1$. Then the value of the expression is $1 + 12 = 13$. Pick another value, say $x=-10$. Then the value is $100-120 = -20$. Notice that $-20$ is less than $13$. You keep picking different values of $x$. The question asks what is the smallest value can you get?

To answer this problem, you need one fact: Smallest value a square can ever be is zero.

Now to use this we can write $$ x^2 + 12 x = x^2 + 12 x +36 - 36 = (x+6)^2 - 36 $$ I have not changed anything. If I put $x=-10$, I get $$ (-4)^2 - 36 = -20$$ same as before. But if I look at $(x+6)^2 -36$, I can't change the $-36$ but to make $(x+6)^2$ as small as possible, I have to set $x=-6$. So the complete answer is

The minimum value of $x^2 + 12 x$ is $-36$ and this happens when $x = -6$.

The trick of writing the expression as a square plus a constant is called completion of squares and you may need it a lot. Here is the formula for this:

If you have $a x^2 + b x$ then add (and subtract) $b^2/(4 a)$ to complete the square.

share|improve this answer
Why the downvote? I find this very helpful to the OP (and the only answer that goes into detail on the "I do not know what is meant by the minimum value" part); wonder why someone gave it a downvote. –  ShreevatsaR Jan 2 at 13:51

The vertex of a parabola is its minimum.

The $x$ coordinate of the vertex can be found by the formula

$$x_v = -\frac{b}{2a}$$ So, \begin{align*} x_v &= -\frac{12}{2(1)} \\ &= -\frac{12}{2} \\ &= -6 \end{align*}

To find the $y$ coordinate of the vertex, substitute $x_v$ into $f(x)$ for $x$.

\begin{align*} y_v = f(x_v) &= x_{v}^{2} + 12x_{v} \\ &= (-6)^{2} + 12(-6) \\ &= 36 -72 \\ &= -36 \end{align*}

So, the minimum is at

$$(-6, -36)$$

share|improve this answer

Usually, when we have to find the minimum of a quadratic expression, we try to complete the square.
We have $y=x^2 + 12x $ . We want to find the minimum possible value of y.
$$y=x^2 + 12x \\= x^2 +2(6)(x) + 6^2 - 6^2 \\=(x+6)^2 - 36 $$
The lowest possible value of the term in square i.e. $(x+6)^2$ is $0$ when $x = -6$.
So the lowest possible value of $y = x^2 + 12x$ is $-36$.
Here is a graph.

Another way to do it is by using the vertex formula, which @okarin has done.

share|improve this answer


Notice that $$y=x^2+12x=(x+6)^2-36$$ represents a parabola.

This parabola has the minimum value at its vertex $(-6,-36)$.

Hence, the answer is $-36$.

share|improve this answer

This is a quadratic equation and can be put in the form of $y=ax^2+bx^2+c$ or $(x^2+12x+0)$. If a is negative the parabola it makes opens downwards. If a is positive like it is in this case, the parabola opens upwards. If it open upwards, it has a minimum or a maximum when opening downwards. So, the vertex is always the minimum or maximum. Minimum for $a>0$ or maximum for $a<0$. So, if you can find the vertex you will have it. As you see it can be rewritten as $(x+6)^2-36$, this is the vertex form of the parabola. The vertex form is a quadratic equation written as such, $a(x-h)^2-k$, where $(h,k)$ is the vertex. So, the minimum is $(-6,36)$.

share|improve this answer
There's no quadratic equation here; just a quadratic expression / function. –  ShreevatsaR Jan 2 at 5:12

In general, to find the minimum of a quadratic function $ y = ax^2 + bx + c $, we can use a few different methods. Note that this function has a minimum iff $ a \ge 0 $. If $ a \le 0 $, it has a maximum, not a minimum.

Complete the square

Here, we try to write the function in the form $ y = a \cdot (x - h)^2 + k $. Once we have this, we know that $ a \cdot \left( x - h \right)^2 \ge 0 $, so $ y \ge k $. We can see this my graphing it as well. The vertex would be at $ (h, k) $.

An example of this would be the function $ y = x^2 + 12x = \left( x + 6 \right)^2 - 36 $, so the minimum value is $-36$, which occurs at $x=-6$.


It is a general rule, which can be proved using the complete-the-square method that the vertex of a parabola with equation $ y = ax^2 + bx + c $ occurs at $ x = - \dfrac {b}{2a} $. Plugging in the $x$-value from here, we can find the $y$-coordinate of the vertex. Another proof of this uses Calculus.

Let $ f(x) = ax^2 + bx + c $. Then, differentiating, we have $ f'(x) = 2ax + b $. Setting this to $0$, we have $$ 2ax + b = 0 \implies x = - \dfrac {b}{2a}. $$

share|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.