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

Question

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

anonymous · Answer

Please help, i really dont get this

calculusxy · Answer

I am working on it.

anonymous · Answer

thanks :)

calculusxy · Answer

I believe the answer is 0

anonymous · Answer

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

calculusxy · Answer

Do you know the numbers to fill the variables?

anonymous · Answer

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 · Answer

I don't know now except for 0

anonymous · Answer

I dont think thats 0 is right

calculusxy · Answer

Sorry about that.

sasogeek · Answer

$\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 · Answer

The very first one

anonymous · Answer

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

anonymous · Answer

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

calculusxy · Answer

Is that a quadratic equation? I believe.

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

my paper says its in standard form

amistre64 · Answer

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

calculusxy · Answer

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+\-_____)