How to find the increasing and decreasing intervals of a quadratic equation without calculus

Question

I want to find the increasing and decreasing intervals of a quadratic equation algebraically without calculus.

The truth is I'm teaching a middle school student and I don't want to use the drawing of the graph to solve this question.

Answer 1

This is if you do not want to use the fact that you know how a graph looks like:

You can explain to him how from this $f(x)=ax^2+bx+c$ arrive at this $f(x)=a(x+ \dfrac{b}{2a})^2 + (c- \dfrac{b^2}{4a})$.

For $x= - \dfrac {b}{2a} + d$ and $x= - \dfrac {b}{2a} - d$ you have the same function value $f(x)$ for whatever $d$ you choose, which means, by this symmetry of$f$, that if function increases when going from the left of $x=- \dfrac {b}{2a}$ towards $x=- \dfrac {b}{2a}$ then it will decrease when going from $x=- \dfrac {b}{2a}$ towards right of $x=- \dfrac {b}{2a}$ and it is either maximal or minimal at $x=- \dfrac {b}{2a}$ (you know that this all depends on whether $a>0$ or $a<0$).

Suppose $a>0$, then, if $x_1>x_2\geq - \dfrac {-b}{2a}$ you obtain $f(x_1)-f(x_2)=a(x_1 + \dfrac{b}{2a})^2 -a(x_2 + \dfrac{b}{2a})^2 =a(x_1-x_2)(x_1+x_2+\dfrac{b}{a})>0$ so $f(x_1)>f(x_2)$. Now use symmetry of the function around $- \dfrac {b}{2a}$ to deduce reverse inequality for the other side.

Similarly for $a<0$.

Answer 2

You could take your quadratic function, call it f and consider the quantity $f(y)-f(x)$ with $y>x$. For example, if $f(x)=x^2$ then this would yield $$y^2-x^2=(y+x)(y-x).$$ This quantity is positive provided $y+x>0$ or $y>-x$. This means $y>|x|$ and in particular $y>0$.

Answer 3

As someone mentioned in the comments, the standard way to do this is the trick of completing the square (also often used to derive the quadratic formula). You just write $$ ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.$$

To see this formula is true, just multiply out the square on the right hand side carefully. You don't need to memorize the trick exactly... just remember vaguely what it looks like and choose the term inside of the parentheses on the right-hand-side to make the $bx$ term come out correctly. Then the $-b^2/4a$ just cancels out the third term from the squaring.

Then use graphing sense to think about the right hand side. It is a parabola of the form $y = ax^2$ (which has vertex at $x=0$) that is shifted to the left by $b/2a$ and up by $c-b^2/4a.$ So the vertex is at $x=-b/2a.$ If $a>0$ it opens upward, otherwise down.

To prove algebraically that $x^2$ is increasing for $x>0$ and decreasing for $x<0$ we can use the fact that $y^2>x^2$ if and only if $|y|>|x|.$ For the function to be increasing on an interval we need $|y|>|x|$ whenever $y>x$ for all $x$ and $y$ in the interval. This is clearly true if both $x$ and $y$ are positive so we must show that the function is not increasing outside $(0,\infty)$. Let $x<0.$ There is no interval around $x$ on which the function is increasing since if we choose some $y$ with $x<y<0$ then $|y|<|x|.$

Exact same logic to show decreasing on $(-\infty,0)$. Other parabolas are just reflections/shifts of this one as completing the square shows.