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 Could anyone help me with finding the vertex of a parabola from an equation? An example equation I have is f(x)=x^(2)-2x+1 [Quote] [Link]
 Well an inverse that is a function doesn't exist but by inverse I meant reverse mapping. I don't work well with words, but if you give me the function I will look at it and see what I come up with. [Quote] [Link]
 The vertex of a parabola will always have an x coordinate of X = -b/2a. Then solve for the y value. a is the coefficient of the x^2 term; b is the coefficient of the x term; c is the constant. [Quote] [Link]
 :D Thanks! That helped a bit. [Quote] [Link]
 Hydrogen777 Administrator5988 Posts11 Cresco 13:0 - 17.18.91821.938 days ago Joshua0317 said:Well an inverse that is a function doesn't exist but by inverse I meant reverse mapping. I don't work well with words, but if you give me the function I will look at it and see what I come up with.It's f(x) = x3 - x. I can't think of a mapping that would work (the mapping would certainly not be a function.) I did however think of a workaround. Since the limit x→∞ = ∞, and the limit x→-∞ = -∞, and since the curve is everywhere differentiable, there's a theorem that states the curve must pass through every y-value between, which means it's onto R. Hopefully the professor won't mind that I didn't prove that f is differentiable. >_> [Quote] [Link]
 My Kumon didn't explain it well so... What's an orthocenter and a circumcenter? I don't really get it... [Quote] [Link]
 Well, all polynomials are differentiable, so they shouldn't mind all that much. [Quote] [Link]
 Hydrogen777 Administrator5988 Posts11 Cresco 13:0 - 17.48.59821.923 days ago True. By the way, if anyone has any idea how to interpret the f([0, 1)) thing (or whatever it was) let me know. I am suspicious of the integral interpretation, and I have no idea how to search for this information on the Internet. [Quote] [Link]
 It looks like an integral for a range, or perhaps coordinates. What I've never seen before is a bracket on the open end and a parenthesis on the close end. That is confusing me. But the last time I did calculus was a half decade ago. I have a friend with a PhD in the maths. I have e-mailed him to see what he has to say. [Quote] [Link]
 Hehe. I will refer back to this thread later today. (yayz for school not) p.s. I thought that the sup was very appropriate. [Quote] [Link]
 Hydrogen777 Administrator5988 Posts11 Cresco 13:0 - 19.98.47821.798 days ago It looks like an integral for a range, or perhaps coordinates. What I've never seen before is a bracket on the open end and a parenthesis on the close end. That is confusing me. But the last time I did calculus was a half decade ago. I have a friend with a PhD in the maths. I have e-mailed him to see what he has to say.Well normally you would exclude a point because the function is not continuous at the point (in which case you'd have to use a limit to evaluate the integral). What's weird is that the function is definitely continuous at ln2, so I see no reason why that point would have been excluded. : / And thank you for emailing your friend. [Quote] [Link]
 Okay I had some coffee so let me try this again. When mapping R to R, to prove onto, you just have to show that all the elements of.the domain map to a value in the range, and that all the values of in the range are used. Basically just show that all y values will be generated. I think that your differential proof is acceptable. [Quote] [Link]
 he. I do have maths homework but it's a poster... [Quote] [Link]
 wolf said:Let me go into math teacher mode now: f([0, ln2)) is a set, and is called the image of the interval [0,ln2) under the function f. It is the set {f(x) such that x is in [0, ln2)}. To generate this set, you go through every point in the interval [0, ln2), apply the function f, and collect the results in the set f([0, ln2)). For f(x)=e^x, it is simple to determine f([0,ln2)) because e^x is continuous and strictly increasing. Strictly increasing means that if 0
 Hydrogen777 Administrator5988 Posts11 Cresco 13:1 - 9.99.31821.298 days ago Of course, that makes a lot of sense. Of course, I turned in the homework like a half hour ago. [Quote] [Link]
 xD. Sorry I couldn't get it to you faster. [Quote] [Link]
 Stupid question: what IS 14x-3? My brain's gone dead XD [Edit:] It's -42 am I right? [Quote] [Link]
 If x = -39/14, then yes. [Quote] [Link]
 erm... okay? It's only year 8 maths, nothing major. [Quote] [Link]
 Okay, a less snarky answer is you can't tell what it is without knowing what x is. [Quote] [Link]
 Unless you mean what is x if 14x-3=0. In that case x would equal 3/14 [Quote] [Link]
 He might have meant 14 * -3 [Quote] [Link]
 Hydrogen777 Administrator5988 Posts11 Cresco 13:1 - 12.88.6821.154 days ago xD. Sorry I couldn't get it to you faster.Haha, no biggie. Thanks for trying, and thank your friend for me if you get the chance. At least I understand it now. [Quote] [Link]
 awesomeguy said:He might have meant 14 * -3 Oh, yes, of course you're right. Please ignore the nonsense I wrote above. [Quote] [Link]
 and in that case, he's right. (seems no one actually confirmed that yet) [Quote] [Link]
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