## anonymous 3 years ago How do you put y=0.05x^2-x+1 in general form Please help i give medals !

1. anonymous

2. calculusxy

I am working on it.

3. anonymous

thanks :)

4. calculusxy

I believe the answer is 0

5. anonymous

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

6. calculusxy

Do you know the numbers to fill the variables?

7. 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

8. calculusxy

I don't know now except for 0

9. anonymous

I dont think thats 0 is right

10. calculusxy

11. 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?

12. anonymous

The very first one

13. anonymous

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

14. anonymous

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

15. calculusxy

Is that a quadratic equation? I believe.

16. anonymous

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

17. amistre64

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

18. anonymous

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

19. amistre64

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

20. anonymous

my paper says its in standard form

21. amistre64

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

22. 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+\-_____)