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
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vertex of parabola $\displaystyle \displaystyle \frac{-b}{2a}$ 2. factor $\displaystyle \displaystyle (3x\pm g)(x\pm f)=0$ where $\displaystyle \displaystyle f,g\in\mathbb{R}$ 3. solve this $\displaystyle \displaystyle 0>9+3x-12x^2$
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 $\displaystyle ax^2+bx+c = 0 , x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Originally Posted by Peyton Sawyer 3.)solve inequality -3x > 9-12x^2 $\displaystyle 12x^2-3x-9 > 0$ Use the same method in Q2 if you like.
i dont understand 2 and 3. can you walk me through them?
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.
for #1 i got -5/12 wat would the vertex be?
I think it should be 1/4
how? i though vertex form was -b/2(a)
b= and a=-10 Thus, -b=-5 and 2a=-20. $\displaystyle \displaystyle \frac{-5}{-20}\rightarrow\frac{1}{4}$ Also note that is only the x coordinate. You must plug it in to obtain the y value.
i didnt rearrange the problem that is why. okay now i plug in the x coordinate into -10x^2-5x+6?
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.
-10(1/4)^2-5(1/4)+6= 5? is that correct?
Nope.
can you help me get y value?
$\displaystyle \displaystyle \frac{-10}{4^2}-\frac{5}{4}+6=\frac{-10}{16}-\frac{20}{16}+\frac{96}{16}=\frac{-10-20+96}{16}=\frac{33}{8}$
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