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- anonymous

How do you put y=0.05x^2-x+1 in general form
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- anonymous

How do you put y=0.05x^2-x+1 in general form
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- katieb

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- anonymous

Please help, i really dont get this

- calculusxy

I am working on it.

- anonymous

thanks :)

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- calculusxy

I believe the answer is 0

- anonymous

It has to be put in general form and its equation is Y=a(x-h)^2+k

- calculusxy

Do you know the numbers to fill the variables?

- anonymous

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

- calculusxy

I don't know now except for 0

- anonymous

I dont think thats 0 is right

- calculusxy

Sorry about that.

- sasogeek

\(\large y=0.05x^{2}-x+1 \)
\(\large y=0.05x^{2-x}+1 \)
\(\large y=0.05x^{2-x+1} \)
which one is your question?

- anonymous

The very first one

- anonymous

y = 0.05x(x-20) + 0.01 , u want in this form?

- anonymous

i dont think so, the eqaution says y=a(x-h)^2+k

- calculusxy

Is that a quadratic equation? I believe.

- anonymous

im not sure, it might be called that or the general form eqaution

- amistre64

general form is: y = ax^2 + bx + c

- anonymous

i keep getting x=.05(x-10)-4 i think imright but im not sure

- amistre64

y = 0.05x^2 - x + 1 , is in general form, unless theres some nitpicking rule about integer coeffs as opposed to ratioanl coeffs

- anonymous

my paper says its in standard form

- amistre64

standard form appears to be:\[y=a\left(x+\frac{b}{2a}\right)^2-\left(\frac{b^2-4ac}{4a}\right)\]

- calculusxy

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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