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

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.

• Why not teach him how to graph it? That would be a good skill to have moving forward. Then you can discuss the change from increasing to decreasing depending on which side of the vertex you are on. – Joel Oct 6 '17 at 3:48
• @Joel He knows how to graph it. I want to teach him how to solve this question algebraically. – user42912 Oct 6 '17 at 3:52
• This is a strange restriction, but them it's your student... Complete the squares to see the max/min (depending on the direction) vertex point. – zipirovich Oct 6 '17 at 3:59
• @zipirovich I think it's more formal to prove it using algebra instead of just drawing the graph. I know how to find the max/min, but I didn't understand what to do with this information (remember I can't use the drawing of the graph) – user42912 Oct 6 '17 at 4:03
• The devising of a method for the solving of quadratic equations should result in a way of finding the coordinates of the associated parabola's vertex... Can you just show him how to solve second-degree equations? – wet Oct 6 '17 at 4:06

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 $a<0$.

• in the eighth line I think you meant $-\frac{b}{2a}$ Thank you very much – user42912 Oct 6 '17 at 13:08
• @user42912 I think not, why would I mean that. All looks well, just the way you requested, hopefully- – user480281 Oct 6 '17 at 14:24

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 $y>0$.

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.

• You use the fact that you know how a graph looks like. – user480281 Oct 6 '17 at 4:23
• Thank you for trying to solve my problem. However the issue with your answer is you use a graph strategy in last paragraph to solve it. I would like to solve this question algebraically. (In the same way we find the max/min of the quadratic equation). – user42912 Oct 6 '17 at 4:23
• @user42912 Well are we not allowed to use translation? In any event we can do it algebraically for $x^2$ and then either by (sensible) translation/reflection or (less sensible) throwing in some b's and a's and cranking the same calculation only way more cumbersome, we can get it for the general case. Have added an algebraic proof for $x^2.$ – spaceisdarkgreen Oct 6 '17 at 4:56