# Thread: Another Increasing/Decreasing Functions Question

1. ## Another Increasing/Decreasing Functions Question

"Use calculus techniques to show that the graph of the quadratic function f(x) = ax^2 + bx + c, a > 0, is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)."

I have no idea where to start really. Can someone lead me in the right direction?

Thanks!

2. Originally Posted by Jeavus
"Use calculus techniques to show that the graph of the quadratic function f(x) = ax^2 + bx + c, a > 0, is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)."

I have no idea where to start really. Can someone lead me in the right direction?

Thanks!
find the derivative and note that it is negative on the interval x < (-b/2a) and positive on the interval x > (-b/2a)

3. f(x) = ax^2 + bx + c
f'(x) = 2ax + b
0 = 2ax + b
-b = 2ax
-b/2a = x

How do I show that it is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)?

4. Originally Posted by Jeavus
f(x) = ax^2 + bx + c
f'(x) = 2ax + b
0 = 2ax + b
-b = 2ax
-b/2a = x

How do I show that it is decreasing on the interval x < (-b/2a) and increasing on the interval x > (-b/2a)?
there is only one point where the derivative is zero, so we don't really have to worry about fluxuations. just pick a point to the right of -b/2a and one to the left. for example, you may consider the derivative at x = -b/2a + 1 and x = -b/2a - 1. see what happens

5. Originally Posted by Jhevon
there is only one point where the derivative is zero, so we don't really have to worry about fluxuations. just pick a point to the right of -b/2a and one to the left. for example, you may consider the derivative at x = -b/2a + 1 and x = -b/2a - 1. see what happens
I know what you're talking about, but I don't know how to represent it on paper. How do I show that it is increasing on the one interval, but decreasing on the other interval?

(I know that the quadratic function looks like a parabola and I can think about how at the zero slope point it would change from decreasing to increasing, but I can't represent it)

6. ## Here you go

$\displaystyle f'(x)=2ax+b$ which equals zero at $\displaystyle x=\frac{-b}{2a}$...now to test before $\displaystyle \frac{-b}{2a}$ put $\displaystyle \frac{-b}{2a}-1$ in the derivative...you get$\displaystyle f'\bigg(\frac{-b}{2a}-1\bigg)=2*a*\bigg(\frac{-b}{2a}-1\bigg)+b=-2a$..assuming that $\displaystyle a>0$ if $\displaystyle a<0$ [tex]f(x)]/math] is decreasing on $\displaystyle x\in\bigg(-\infty,\frac{-b}{2a}\bigg)$

QED

7. Ahh, okay. After plugging in (-b/2a - 1) and (-b/2a +1) into the derivative function, I get -2a and 2a respectively.

Meaning that the function is decreasing on the interval x < -b/2a and the function is increasing on the interval x > -b/2a.

Thank you.

8. Originally Posted by Jeavus
Ahh, okay. After plugging in (-b/2a - 1) and (-b/2a +1) into the derivative function, I get -2a and 2a respectively.

Meaning that the function is decreasing on the interval x < -b/2a and the function is increasing on the interval x > -b/2a.

Thank you.
good. and note that you know that -2a is negative and 2a is positive respectively because they told you a > 0. you have to watch out for that