## anonymous one year ago Help with this parabola, 3x^2 +4x+12

1. anonymous

@Nnesha

2. campbell_st

well I can tell you its positive definite

3. Nnesha

statement ? factor ? solve for x ? graph ? write in vertex form? find vertex point ? what is axis sym ?

4. anonymous

I dont know how to find the zeros

5. Nnesha

alright zeros(solutions , x-intercept) you can use quadratic formula OR you can factor it

6. Nnesha

which one is easy for you ? factors or quadratic formula ?

7. anonymous

Factors

8. Nnesha

ohh well nvm you have to apply the quadratic formula $\huge\rm x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$

9. Nnesha

ax^2+bx+c quadratic equation where a=leading coefficient b=middle term c-constant term so what is a b and c in that equation ?

10. Nnesha

$$\huge\color{reD}{\rm b^2-4ac}$$ Discriminant you can use this to find if the equation is factorable or not if  b^2-4ac > 0 then there are 2 real zeros if  b^2-4ac = 0 then there is one real root if  b^2-4ac < 0 then you will get two complex roots (no -x-intercept)

11. Nnesha

zeros is another word they use for solutions and x-intercept

12. LynFran

can u show me how to write it in vertex form .. @Nnesha

13. Nnesha

complete the square y=a(x-h)^2+k ?

14. LynFran

o thats all lol... im now doing this ..thanks

15. Nnesha

i mean convert 3x^2 +4x+12 into y=a(x-h)^2+k <---(vertex form) b using completing the square method :=)

16. LynFran

i know what u meant thanks

17. Nnesha

np :)

18. LynFran

so it would be 3x^2+(4x/2)^2+......+12-..... (3x^2+4x+4)+12-4 (3x+2)^2+8 is this correct

19. LynFran

i think i did something wrong

20. Nnesha

okay you need to complete the square of (3x^2+4x) it would be great if you take out the 3

21. Nnesha

i've to go rn cya later

22. LynFran

ok