Help! Algebra!!!

- anonymous

Help! Algebra!!!

- katieb

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- anonymous

What is the domain of the following parabola?

##### 1 Attachment

- anonymous

- anonymous

the domain is all x values @MafHater

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## More answers

- Keigh2015

Is there any choices? and yeah the domain is all the x values

- anonymous

Are there

- anonymous

Yes,
x ≥ 1
x ≤ 1
All real numbers
y ≤ 2

- anonymous

I know it's not D, I think it's A.

- anonymous

all real numbers

- Keigh2015

what do you think it is? @MafHater

- anonymous

bc it keeps extending through all numbers of x

- anonymous

Like I said @Keigh2015, I believe it is A. Is @DaBest21 right?

- anonymous

as u can see it is not a bc it goes greater than one

- anonymous

so it hjas to be all real numbers

- Keigh2015

I think @DaBest21 is correct, because yeah the x-values go farther out than one.

- anonymous

Don't forget to medal and fan if you already haven't. Plus send me a message if you need more help.

- anonymous

Okay, cool! Thanks!
Would you mind helping me on a few more?

- Keigh2015

I wouldn't mind helping

- anonymous

The quadratic functions f(x) and g(x) are described as follows:
f(x) = −4x2 + 5
x g(x)
0 0
1 1
2 5
3 1
4 0

- anonymous

Which of the following statements best compares the maximum value of the 2 functions?
It is the same for both functions.
f(x) has a greater maximum value than g(x).
g(x) has a greater maximum value than f(x).
The maximum values cannot be determined.

- anonymous

sure just make a new question in ask a question

- anonymous

Is it C?

- anonymous

at what point in the chart is g(x) at its max

- anonymous

g(x) is at its max in the middle, f(x) is at its max at the end.

- anonymous

what amounts

- anonymous

Hint:they are equal

- Keigh2015

https://www.desmos.com/calculator/yrpnondjwl this is what it would look like graphed

- anonymous

because the greater the number or less the less it is so the both max at 5 so they are equal

- anonymous

it is A @MafHater

- anonymous

Ohh, ok. @Keigh2015 what do you think?

- anonymous

send me a message if you need more help.

- Keigh2015

I agree with @DaBest21

- anonymous

Cool, thanks!

- Keigh2015

No problem

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