# Math Help - Matching the functons with their graphs

1. ## Matching the functons with their graphs

How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.

Any help would be greatly appreciated!

2. wouldnt just using a graping calculator and inputing it in make life sooo much easier?

3. answer to number 10 would be
1 D
2 C
3 B
4 A

4. Originally Posted by s3a
How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.

Any help would be greatly appreciated!
What you SHOULD be doing is showing us your work, so that we can show you where you're going wrong.

Having said that...

7) $3x^4 - 3x^3 - 1x^2 = 0$

$x^2(3x^2 - 3x - 1) = 0$

So $x^2 = 0$ or $3x^2 - 3x - 1 = 0$

Therefore

$x = 0$ or $x = \frac{3 \pm \sqrt{(-3)^2 - 4(3)(-1)}}{2(3)}$

$x = 0$ or $x = \frac{3 \pm \sqrt{13}}{6}$.

So $A = \frac{3 - \sqrt{13}}{6}, B = 0, C = \frac{3 + \sqrt{13}}{6}$.

5. Originally Posted by s3a
How do I do this? If someone could at least show me how to do one of them with full solutions, it would help me understand them all.

Any help would be greatly appreciated!
8) $5 - x < -1$

$5 + 1 < x$

$x > 6$.

Using interval notation, this is written as $x \in (6, \infty)$.

9) $x^3 - 16x \leq 0$

The easiest way to answer this would be to analyse the graph of $y = x^3 - 16x$. See where the x-intercepts are and where the graph is negative.

6. Actually, I needed detailed solutions for number 10 (or one of the graphs), sorry I didn't make that clear; I was in a rush.

7. Originally Posted by saar4ever
wouldnt just using a graping calculator and inputing it in make life sooo much easier?
Of course it would. And having the answers ahead of time would save the trouble of using a calculator! You wouldn't learn anything in either case, but that's not important is it?

8. Originally Posted by s3a
Actually, I needed detailed solutions for number 10 (or one of the graphs), sorry I didn't make that clear; I was in a rush.
Well, what do you KNOW about functions? Can I presume that you know how to factor?

Function (1), $f(x)= -x^2(x^2- 4)= -x^2(x-2)(x+2)$ obviously has zeros at x= 0, 2, and -2 and only at those points. Only graphs A and D have that property. For x> 2, all factors, x, x-2, and x+2, are positive so, because of the "-" in front, the f(x)< 0 for x> 2 and only D has that property.

Function (2), $f(x)= -x^5+ 5x^3- 4x= -x(x^4- 5x^2+ 4)$ $= -x(x^2- 4)(x^2- 1)= -x(x-2)(x+2)(x-1)(x+1)$ has zeros at x= 0, x= 2, x= -2, x= 1, and x= -1. C is the only graph that has that property.

Function (3), $f(x)= -x^3+ 2x^2= -x^2(x- 2)$ has zeros only at x= 0 and x= 2 (and a double root at x= 0 so the graph is tangent to the x-axis at x= 0). B is the only graph having that property.

Function (4), $f(x)= \frac{x^6}{2}- 2x^4=\frac{1}{2}x^4(x^2- 4)= \frac{1}{2}x^4(x-2)(x+2)$ has zeros at x= 0, x= 2, and x= -2 and, as with (1), A and D have that property. But now, for x> 2, f(x)> 0 and only A has that property.

9. I was reading and in the beginning I noticed that (x+2) or (x-2) is negative (in my work - as opposed to positive in yours) (can't tell which one is negative and which one is positive).