## csimmons3 Group Title How do you put y=0.05x^2-x+1 in general form Please help i give medals ! one year ago one year ago

1. csimmons3 Group Title

2. calculusxy Group Title

I am working on it.

3. csimmons3 Group Title

thanks :)

4. calculusxy Group Title

I believe the answer is 0

5. csimmons3 Group Title

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

6. calculusxy Group Title

Do you know the numbers to fill the variables?

7. csimmons3 Group Title

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 Group Title

I don't know now except for 0

9. csimmons3 Group Title

I dont think thats 0 is right

10. calculusxy Group Title

11. sasogeek Group Title

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

12. csimmons3 Group Title

The very first one

13. tanjeetsarkar96 Group Title

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

14. csimmons3 Group Title

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

15. calculusxy Group Title

Is that a quadratic equation? I believe.

16. csimmons3 Group Title

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

17. amistre64 Group Title

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

18. csimmons3 Group Title

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

19. amistre64 Group Title

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

20. csimmons3 Group Title

my paper says its in standard form

21. amistre64 Group Title

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

22. calculusxy Group Title

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+\-_____)