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How do I find the domain of this function?

Mathematics
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\[g(x) \sqrt{x^2 - 3x - 40}\]
I have \[\sqrt{(x-8)(x+5)} \]
\[(x-8)(x+5) \ge 0\]

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Other answers:

The original problem is at the top.
Draw a number line and see for it
I don't know what x can't be though
I know the answer but I don't know how to get the answer.
(-infinity , 5] U [8 , infinity)
yeah but how do I find that?
just do trial and error method...)
(x−8)(x+5)≥0
luk for the vlue of x which satisfies this...
There must be a logical way to find the answer, right?
(x-8)(x+5) >= 0 its a parabola, right ?
I could input it into my graphing calculator but other wise I don't know how to tell if its a parabola or not.
y = ax^2 + bx + c is a parabola
any quadratic function gives a parabola graph
ok
can you find the x intercepts ?
once you find x-intercepts, you would know where the parabola dips into x axis (goes negative)
|dw:1346519548989:dw|
Oh ya that would be the point at which the g(x) function is 0, so its x - 8 = 0, x = 8, x+ 5 = 0, x=-5, and since its a parabola the graph only passes through zero twice. Nice dude, so the domain is (-infinity, -5] U [8, +infinity]
yeah nice you got it :)
Awesome, thanks again. Is there a way I could recognize any graph's shape or at least some and would you recommend memorizing those?
if the degree is 2, then its a parabola
finding domains for other polynomials is bit tricky... as they may dip into x-axis multiple times
okay
do you knw sketching polynomilas (end behiavior thingy )
never heard of it
end behavior helps you sketch ANY polynomial freehand
Thanks dude, I'll look it up
process is basically : 1) find x intercepts 2) using end-behavior sketch the graph by hand
its easy concept and worth learning... im sure you wil get hang of it in few minutes....... good luck :)
why is it that a polynomial with a degree of two is a parabola?
Thanks again.
uhh il have to think...... idk exactly... i think it has something to do with local maximum/minimum + increasing/decreasing thingy. we need to turn to calculus for proper understanding. im also learning... . so im no good for answering this :(
Oh, I have enough to learn for now so it doesn't matter to me. I bet I wouldn't understand the explanation if it involves terms I've never had before haha.

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