Question

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

Answer 1

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

Answer 2

the domain of all polynomial function is all real numbers.

Answer 3

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

Answer 4

:)

Answer 5

But what is the range?

Answer 6

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

Answer 7

You still there, @Photon336 ?

Answer 8

@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.

Answer 9

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. :)

Answer 10

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

Answer 11

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

Answer 12

so for a parabola the general formula is this:

\[ax ^{2}+bx+c\]

Answer 13

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

Answer 14

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

right?

Answer 15

look at this

Answer 16

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. ???

Answer 17

@hhopke Domain = all x values

Answer 18

while range is all y values

Answer 19

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)

Answer 20

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

Answer 21

So what am I doing wrong? :(

Answer 22

@ganeshie8

Answer 23

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

Answer 24

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. :(

Answer 25

@ganeshie8

Answer 26

This happened with all the problems, if that helps.

Answer 27

\[3x^2+6x-1\\~\\=3(x^2-2x) - 1 \\~\\= 3(x^2-2x+1)-4 \\~\\= 3(x-1)^2-4\\~\\\ge -4\]

therefore the range is \([-4, \infty)\)

Answer 28

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

Answer 29

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

Answer 30

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