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

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

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

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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.

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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

$$(-6, -36)$$

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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.

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HINT :

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 $-36$.

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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 $(-6,36)$.

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There's no quadratic equation here; just a quadratic expression / function. – ShreevatsaR Jan 2 '14 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 $x=-6$.

Formula

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 $$2ax + b = 0 \implies x = - \dfrac {b}{2a}.$$

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