What is the range of function 3x^2+6x-1 in interval notation?

- hhopke

What is the range of function 3x^2+6x-1 in interval notation?

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- hhopke

Also, according to the HW program, \[(-4,\infty)\] is wrong.

- anonymous

the domain of all polynomial function is all real numbers.

- hhopke

Yeah. Domain is \[(-\infty,\infty)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

:)

- hhopke

But what is the range?

- hhopke

If you plug it into a graphing calculator, it looks like this:

##### 1 Attachment

- hhopke

You still there, @Photon336 ?

- Photon336

@hhopke
since it's quadratic
you could use part of the quadratic formula, and see what value it is whether if it's greater than zero, this will give you a sense of it's real or imaginary. i feel that this is important because if b^2-4ac < 0 then your function is going to be imaginary, so i guess not all real numbers will apply to this.
\[b ^{2}-4ac \] > 0
b = 6
a = 3
c = -1
(6)^2-4(3)(-1) = 48
it's all real numbers, it's going to be continuous everywhere, every x value has a distinct y v value.
my math skills are weak, but usually if a function is continuous everywhere it's also differentiable on that particular interval. if we take the derivative.
\[\frac{ dy }{ dx } 3x ^{2}+6x-1 = 6x+6 \] that's continuous for x value.

- hhopke

Thank you, but I have absolutely no clue what you just said. We just started this. All the information I have on the subject (and supposedly, all I need) is that if a parabola concaves up, then the range is anything greater than the Y value of the vertex, and if it concaves down, then the range is anything less than the Y value of the vertex. Really basic stuff- no quadratic formula yet. :)

- Photon336

@hhopke yeah, sorry about that lol I thought you were asking for the range

- hhopke

I was asking for range. Everything else I search on the internet gives me all sorts of weird stuff using set notation and such... :(

- Photon336

so for a parabola the general formula is this:
\[ax ^{2}+bx+c\]

- Photon336

@hhopke the sign in front of the a value tells us whether a prabola will concave up or down

- hhopke

Right. This one is up, b/c 3>0
right?

- Photon336

look at this
|dw:1444541020427:dw|

- hhopke

Yeah. When I graphed it on my calculator, it gave me a concave-up parabola. The vertex was at (-1,-4). But when I tried to do the range in interval notation, I typed (-4,oo) and the program said it was wrong. ???

- hhopke

##### 1 Attachment

- Photon336

@hhopke Domain = all x values

- Photon336

while range is all y values

- hhopke

All possible Y-values for the function is the range of that function. The lowest possible y-value for this particular function is -4. Interval notation for the range should be (-4, oo) right? (oo is infinity)

- Photon336

yeah I just plotted your graph on wolfram alpha. -4 is the lowest y value

- hhopke

So what am I doing wrong? :(

- Photon336

- Photon336

i'm not sure, let me tag someone who knows better

- hhopke

Alrighty. Same thing happened w/ function 3x^2. Plugged it in, lowest y-value was 0. Plugged in (0, oo) for range and it was marked wrong. :(

- hhopke

- hhopke

This happened with all the problems, if that helps.

##### 1 Attachment

- ganeshie8

\[3x^2+6x-1\\~\\=3(x^2-2x) - 1 \\~\\= 3(x^2-2x+1)-4 \\~\\= 3(x-1)^2-4\\~\\\ge -4\]
therefore the range is \([-4, \infty)\)

- hhopke

ooooooooooooooohhhhhhhhhhhhhhh.... I formatted it wrong. Thank you sosososososososososososososo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

- ganeshie8

change the left parenthesis, "\((\)", to square backet "\([\)" and you will be okay

- ganeshie8

np, other than that ur work looks good ! nice job!

- hhopke

THANK YOU SO MUCH TO BOTH @ganeshie8 and @Photon336

- Photon336

in the third line, how did you get -4 @ganeshie8 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.