.'.'.' i will give medal '.'.'.
\[y=x^2-2x+1\] Factorise the right hand side. Hence find the two values of x which makes y=0. Write down the y-intercept. Sketch the curve. Write down the equation of the line of symmetry.
ok so step 1, factor. do you know x factoring?
i mean this one you can kind of just visualize the answer quickly but if you need help i can explain it to you
when i factorised it im getting x=1 and x=1 and y-intercept=1 so how to plot it on graph
(x-1)^2 is your factor so yes you got that correct
please explain it from 1 step
on how to factor it? or how to graph it?
If you complete the square it would be easier to graph.
both
So what you want to do is convert general form (what you have now) into standard form. But first, lets do factoring.
Do you know how to factor?
yes
alright, so i like to factor using the x method and if you dont know it then i can explain.first draw an x. then, in your type of equation where you have a variable squared, plus of minus another number times the variable, plus or minus a number, then you take the last number in your equation and put it on the bottom. and you take your middle number and put it on the top of the x
so for you that would look like -2 X 1
then on the two sides find the numbers that add to make the top number and multiply to make the bottom number
of course you can do this the other way around and it will work just the same so 1 X -2
in your case those numbers are -1 and -1. they add to make -2 and multiply to make 1
from there put your equation into the factored form so (x-1)(x-1). the two values inside the parentheses are the same in your case so you can shorten this even more by just writing (x-1)^2 because it means the same thing
with me so far?
if im doing from quadratic formula im getting both x values +1
quadratic formula will get you the same thing so yes that works too
its just my personal preference to use the x method
but you understand the factoring part right?
so how can we plot x=1 and x=1 on graph
alright so this is just a sort of hint i guess but any time you have a variable squared in your equation that means that there will be a parabola.
|dw:1403779008569:dw|
so to make this easy lets just go through this step by step. first off just look at the y=x^2 part you have a standard parabola that has a vertex of (0,0)
what is parabola i've not studied it yet
|dw:1403779130712:dw| horrible drawing on my part but like that
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