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How do you put y=0.05x^2-x+1 in general form Please help i give medals !

Mathematics
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Please help, i really dont get this
I am working on it.
thanks :)

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Other answers:

I believe the answer is 0
It has to be put in general form and its equation is Y=a(x-h)^2+k
Do you know the numbers to fill the variables?
ive gotten so far: y=0.05x^2-x+1 y-1=.05x^2-x y-1+.25=.05x^2-x+.25 y-.75=.05x^2-x+.25
I don't know now except for 0
I dont think thats 0 is right
Sorry about that.
\(\large y=0.05x^{2}-x+1 \) \(\large y=0.05x^{2-x}+1 \) \(\large y=0.05x^{2-x+1} \) which one is your question?
The very first one
y = 0.05x(x-20) + 0.01 , u want in this form?
i dont think so, the eqaution says y=a(x-h)^2+k
Is that a quadratic equation? I believe.
im not sure, it might be called that or the general form eqaution
general form is: y = ax^2 + bx + c
i keep getting x=.05(x-10)-4 i think imright but im not sure
y = 0.05x^2 - x + 1 , is in general form, unless theres some nitpicking rule about integer coeffs as opposed to ratioanl coeffs
my paper says its in standard form
standard form appears to be:\[y=a\left(x+\frac{b}{2a}\right)^2-\left(\frac{b^2-4ac}{4a}\right)\]
I believe that you should find the sum that equals to 1 and the product that's equal to 0.05. However, both the numbers used will have to be same used in the 0.05 and 1. So it can be like (x+\-___)(x+\-_____)

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