Help drawing a function, finding its zero, local max/min

[gigidy-goo]

Hello to all.
I'm having a few problems and would love to know how to do the following.
1. f(x) = 3x^2-8



2. Find the function's zeros, local max/min and the function's behaviour



3. My attempt at drawing the function ended with a downward curve intercepting at y=-8, x=0

But a graphing calculator showed an upward 'smile'.

Thanks in advance for any help. All I want to know is how to find the things above but the graphing calculator threw me off a bit!

 

[armolinasf]

the graph should be upward sloping since 3x^2 is positive, your intercepts are correct though.

The functions zeros are where ever the function is equal to zero so you would solve for 0=3x^2-8 => 8=3x^2

Local min's or max's is given by evaluating the function at -b/2a, in your case a=3, b=0, and c=-8. So the max or min is the point (-b/2a, f(-b/2a)) This just comes from completing the square.

 

[HallsofIvy]

"Ended" at (0, -8)? Did you plot any points with x negative?

 

[gigidy-goo]

Hi armolinasf
Thanks for replying
The slope I drew came from changing x to -2,-1,0,1,2,3,etc so I ended up with a line starting at,
x=-3,y=-26
x=-2,y=-20
x=-1,y=-14
x=0,y=-8
x=1,y=-1
x=2,y=4
x=3, y=10.
x=4,y=16

It is going upward you're right but it doesn't look like the graph from the calculator. Do you know where I went wrong there? The calculator showed an upward smile whereas mine only starts low and goes up.

 

[gigidy-goo]

Hi hallsofivy,

Thanks for replying.

I just meant my attempt left me with a situation where x=0,y=-8.

But I'm not sure why my line doesn't look like an upward slope as per the graph?

 

[eumyang]

The graph of y = 3x² - 8 is not a "line." It is called a parabola.

What are these y-values? If they are supposed to be the y-values in the function
y = 3x² - 8
then they are all wrong, except for x = 0, y = -8. For instance,
x = -1, y = -5
x = 0, y = -8
x = 1, y = -5

If the y-values you listed are NOT what I'm thinking of, what are they, exactly?

 

[eumyang]

I think I can guess where you're getting those y-values. Except for x = 1, y = -1, the other points fit on the line

y = 3 \cdot x \cdot 2 - 8
where you are MULTIPLYING x by 2, whereas the function you originally posted is

y = 3 \cdot x^2 - 8
where you are supposed to SQUARE x before multiplying by 3.

 

[gigidy-goo]

I think I'm way off here. How did you get x = -1, y = -5?

I've tried it a few different ways and end up with different answers;

For x = -1

3(-1^2)-8
-1^2(3)-8

This is probably quite simple stuff but if I know how to do this I can get the rest

 

[gigidy-goo]

Ok so by squaring x before, I'm still not getting the right numbers...

-1^2 = -1
3.-1=-3
-3-8=-11

Where are I going wrong?

 

[Mentallic]

No, if you have x² where x=-1, then the answer is 1. (-1)²=1 which is what you should be doing, while you were doing -1² which you would be doing if you had -x² and had to solve where x=1. Basically that's the same as -(1)²

For x=-2, you would have (-2)²=4 etc.

 

[gigidy-goo]

Ah, right well that's a tad embarrassing.

I have it now so thanks to everyone for helping out, I squared 'x' inside the bracket instead of outside.

That's great so there is only a local min and not a max?

And the behaviour of this function increases as it tends towards infinity, would that be correct?

Thanks again for your help

 

[Mark44]

I think I'm way off here. How did you get x = -1, y = -5?

I've tried it a few different ways and end up with different answers;

For x = -1

3(-1^2)-8
-1^2(3)-8

This is probably quite simple stuff but if I know how to do this I can get the rest

Try putting in = between expressions that have the same value.

For the above, if x = -1, y = 3(-1)^2 - 8 = 3(1) - 8 = -5.

Ok so by squaring x before, I'm still not getting the right numbers...

-1^2 = -1
3.-1=-3
-3-8=-11

Where are I going wrong?

It's really (-1)², which is the same as (-1)(-1) = + 1. Instead of taking baby steps like the above, work with the whole expression you're trying to evaluate, as I did above.

Ah, right well that's a tad embarrassing.

I have it now so thanks to everyone for helping out, I squared 'x' inside the bracket instead of outside.

That's great so there is only a local min and not a max?

And the behaviour of this function increases as it tends towards infinity, would that be correct?

There is a local minimum that also happens to be the global minimum. It's at the low point of the parabola. There is no maximum -- local or global.

Try to be more specific than "as it tends towards infinity..." Anyone reading this would assume that the antecedent of "it" (i.e., the word that "it" represents) was "this function" but I'm pretty sure that's not what you meant. A better way to say this is: The function values increase as x tends to infinity. The function values also increase as x tends to negative infinity.