I need to verify I have the correct vertex of the following.. f(x)=x^2-x+1 I also am unable to figure out how to get the values from the graphing calculator. Anyone know please?

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I need to verify I have the correct vertex of the following.. f(x)=x^2-x+1 I also am unable to figure out how to get the values from the graphing calculator. Anyone know please?

Mathematics
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to find the vertex you'll have to take x = -b/2a then find f(-b/2a) :)
so , I have the following: a = 1, b = -1 and x = 1 \[x = \frac{1}{2}\] and \[f(\frac{1}{2}) = (\frac{1}{2})^2 -\frac{1}{2} + 1 = \frac{3}{4}\] so your vertex is : \[(\frac{1}{2}, \frac{3}{4})\] correct me if I'm wrong ^_^
Wow, that is what I JUST got when I zoomed in my calculator.

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That is wonderful. But on that calculator, how do I find the value?
hmm, I'm not sure, I've never used a graphing calculator before :) I just simply sketch ^_^
why don't you sketch it?
Oh my goodness, thank you . I have been trying to find the vertex formula.
I did sketch it..
most welcome :)
then you can easily plot the vertec from your sketch :)
vertex*
Its asks for values such as XMIN, XMAX, XSCL, YMIN, YMAX AND YSCL
the question? or the graphing calculator?
IIs on the calculator, but it never changes from -10 or 10..
those values should show on the calculator..but i have no clue how to find them and it's part of my answer the teacher wants
lol, I wish I can get my sister here, she uses a graphing calculator , she'll prolly know the answer, but school lol
I'm sorry :(
LOL. Thank you for your help. I will figure it out tomorrow..thanks a bunch

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