I am being ask to explain in two ways why is it that $y=ax^2+bx+c$ parabola opens up if $a$ is positive and why is it that $y=ax^2+bx+c$ opens down when $a$ is negative. One of the explanations has to be understood by beginning algebra student. I am unsure how I would explain it
$\endgroup$ 25 Answers
$\begingroup$If $x$ is big and positive, and $a$ is positive, then $ax^2$ will be very big and positive, overwhelming any effect from $bx+c$.
If $x$ is big and negative, and $a$ is positive, then $ax^2$ will again be very big and positive.
So if $a$ is positive, the parabola opens upwards.
If $a$ is negative then if $x$ is big (positive or negative) the opposite occurs, and $ax^2$ will be very big and negative with the parabola opening downwards.
$\endgroup$ 1 $\begingroup$Hint: Show that the parabola has a unique minima or maxima depending on whether $a$ is positive or negative respectively using the second derivative idea. This is the calculus way. In the algebraic way, I guess plotting is one option.
$\endgroup$ 1 $\begingroup$One more method:
You know the parabola $y = ax^2 + bx+c$ can be written as: $$y - y_0 = a(x-x_0)^2$$ Since this is just the parabola "shifted" to the axis. Now since $(x-x_0)^2$ is always positive, what determines what the parabola looks like is only $a$.
$\endgroup$ $\begingroup$So I am gonna try to prove from "slope" perspective. I hope it's clear that if our parabola opens upward then slope is positive else negative for say large positive x values.
Consider $ y = ax^2 + bx + c $ with 2 points $(x_1,y_1)$ and $(x_2,y_2)$. Also we will safely assume $x_2 > x_1 > 0$ i.e. points are on the right side of Y-axis.
Now, slope = $ \frac{y_2 - y_1}{x_2-x_1} $ = $a(x_1+x_2)+b$.
It must be clear that:
- $b>0$, then it's pretty straightforward that slope will be positive if $a>0$ and hence parabola opens upward else $a<0$ would imply slope negative and parabola opens downward.
- $b<0$, If $a<0$ that implies slope negative and parabola opens downward but if $a>0$ we will get $x_1$ and $x_2$ such that $a(x_1+x_2) > b$ and hence slope positive implying parabola opening upward.
This might convince beginning algebra student else the answer given by Gautam Shenoy about using second-derivative is more concrete.
$\endgroup$ $\begingroup$My usual approach when teaching this topic is a "transformational" approach. We first understand basic transformations: if we know what the graph of $f(x)$ looks like, then
- the graph of $f(x-h)$ is the same graph, but translated to the left by $h$ units;
- the graph of $f(x) + k$ is the same graph, but translated up by $k$ units;
- for positive $A$, the graph of $A f(x)$ is the same graph, but scaled vertically by a factor of $A$; and
- the graph of $-f(x)$ is the same graph, reflected vertically across the $x$-axis.
We also discuss horizontal scaling and reflection, but those are not relevant here. These kinds of transformations are the bread-and-butter of a beginning algebra curriculum, so I would think that they should be pretty accessible at that level.
Next, think about the ur-parabola, $p(x) = x^2$. The graph of this function is a U-shaped beast which opens up. Even better, it is possible to show that the graph of every quadratic polynomial is a transformed version of the ur-parabola. That is:
Proposition: If $a,b,c\in \mathbb{R}$ with $a\ne 0$ (that is, if $a$, $b$, and $c$ are any real numbers with $a$ nonzero), then there are real numbers $A$, $h$, and $k$ such that $$ ax^2 + bx + c = A(x-h)^2 + k. $$
Proof: We will show that such real numbers exist by explicitly solving for them. Observe that $$ \color{red}{a}x^2 + \color{purple}{b}x + \color{blue}{c} = A(x-h)^2 + k = \color{red}{A}x^2 + \color{purple}{(-2Ah)}x + \color{blue}{(Ah^2 + k)} $$ if and only if the coefficients on the left and right agree. That is $$\begin{cases} \color{red}{a = A} \\ \color{purple}{b = -2Ah} \\ \color{blue}{c = Ah^2 + k} \\ \end{cases}$$ Solving for $A$, $h$, and $k$, we obtain $$ A = a, \qquad h = -\frac{b}{2a}, \qquad\text{and}\ \qquad k = c - \frac{b^2}{4a}. $$ As long as $a \ne 0$ (which was assumed at the beginning), then $A$, $h$, and $k$ can be explicitly found, which gives the desired result.$\tag*{$\square$}$
Remark: As a minor aside, notice that we have just derived the "completing the square" formula, and are very close to a complete derivation of the quadratic formula. Thus this form of the parabola has other uses (including several that I haven't listed).
We have just shown that every parabola can be written in the form $$ ax^2 + bx + c = a\left(x - h \right)^2 + +k. $$ In particular, keeping our transformations in mind, we can graph any parabola by translating the ur-parabola to the right $h$ units, scaling it by a factor of $a$, possibly reflecting it vertically if $a < 0$, and translating it up $k$ units (where $h$ and $k$ are constants that we can write down explicitly in terms of $a$, $b$, and $c$). Note in bold that if $a < 0$, then there is a vertical reflection. This finally leads to the desired explanation:
Conclusion: Suppose that we want to know if the graph $y = ax^2 + bx + c$ opens up or down. To answer this question, we can write the original parabola in terms of transformations, i.e. write $$ ax^2 + bx + c = a(x-h)^2 + k, $$ and observe that only transformation that can change the direction in which the parabola opens is the possible vertical reflection that occurs if $a < 0$. Therefore if $a > 0$ then the parabola retains its original orientation and opens up, and if $a < 0$ then the parabola is reflected vertically and opens down.
$\endgroup$
