The minimum value of $f(x) = 3x^2 + 12x + 180$

Question

I have this equation where the question says:

What is the smallest value, to 2 decimal places, in the image set of the function$$f(x)=ax^2+bx+c,$$where $a = 3$, $b =12$, and $c =180$?

So I used the following formula to get this answer $189.67$:$$\frac{4ac−b^2}{4a}.$$

Then I simply substituted the values. Is that correct?

Answer 1

$f(x)=ax^2+bx+c=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}≥c-\frac{b^2}{4a}$ 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, $ax^2+bx+c=y≥c-\frac{b^2}{4a}$

Answer 2

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.

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$\textbf{Hint}$ : 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.

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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.

Answer 3

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

$$a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$$

You get there by Completing the square.

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

$$\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)$$

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