1. ## Math 111 Help

Ive tried but these are the ones that are giving me trouble....

1.) write in vertex form. Identify the vertex
g(x)=6+5x-10x^2

2.)solve quadratic equation with any method
3x^2-6x-2=0

3.)solve inequality
-3x > 9-12x^2

2. vertex of parabola $\displaystyle \frac{-b}{2a}$

2. factor $\displaystyle (3x\pm g)(x\pm f)=0$ where $\displaystyle f,g\in\mathbb{R}$

3. solve this $\displaystyle 0>9+3x-12x^2$

3. Originally Posted by Peyton Sawyer
1.) write in vertex form. Identify the vertex
g(x)=6+5x-10x^2
do you know how to complete the square?

Originally Posted by Peyton Sawyer
2.)solve quadratic equation with any method
3x^2-6x-2=0

for $ax^2+bx+c = 0 , x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Originally Posted by Peyton Sawyer
3.)solve inequality
-3x > 9-12x^2
$12x^2-3x-9 > 0$

Use the same method in Q2 if you like.

4. i dont understand 2 and 3.
can you walk me through them?

5. For number two, use what pickleslide suggested the quadratic since it isn't factorable.

3 factor out a 3(4x^2-x-3)>0 divide by 3 4x^2-x-3>0 and this can be factored.

6. for #1
i got -5/12
wat would the vertex be?

7. I think it should be 1/4

8. how? i though vertex form was -b/2(a)

9. b= and a=-10 Thus, -b=-5 and 2a=-20. $\displaystyle \frac{-5}{-20}\rightarrow\frac{1}{4}$

Also note that is only the x coordinate. You must plug it in to obtain the y value.

10. i didnt rearrange the problem that is why.
okay now i plug in the x coordinate into -10x^2-5x+6?

11. Yes. You don't need to rearrange always remember it goes ax^2+bx+c. a is with the squared term and b the x.

12. -10(1/4)^2-5(1/4)+6=

5?

is that correct?

13. Nope.

14. can you help me get y value?

15. $\displaystyle \frac{-10}{4^2}-\frac{5}{4}+6=\frac{-10}{16}-\frac{20}{16}+\frac{96}{16}=\frac{-10-20+96}{16}=\frac{33}{8}$