Quadratic equations

[megadeth95]

Hello everyone, I need help with these problems

1) g(x)=6+6x-3x² find the vertex, domain, range and any zeros of the function

2) solve x^-2 -2x^-1 -1 = 0

This is what I did so far

1) the vertex is (1,9), the range is y<9
I forgot how to find the domain and I can't find the other zeros. Don't you have to have one zero in order to find the other ones?

2) 1/x² - 2/x - 1 = 0

1 -2x - x² / x² = 0

1-2x-x² = 0 I get stuck here. Any suggestions?

Thanks for the help!

 

[wjm11]

1) g(x)=6+6x-3x² find the vertex, domain, range and any zeros of the function

2) solve x^-2 -2x^-1 -1 = 0

This is what I did so far

1) the vertex is (1,9), the range is y<9
I forgot how to find the domain and I can't find the other zeros. Don't you have to have one zero in order to find the other ones?

Can you graph problem 1? The zeroes are where the parabola crosses the x-axis. You can find these values by setting the function equal to 0 and solve using the quadratic equation. The domain is simply all the allowable x values. Is there any value that x is not allowed to be?

 

[lookagain]

Hello everyone, I need help with these problems

1) g(x)=6+6x-3x² find the vertex, domain, range and any zeros of the function

2) solve x^-2 -2x^-1 -1 = 0

This is what I did so far

1) the vertex is (1,9), the range is y < 9 No, the range is y <= 9.

I forgot how to find the domain and I can't find the other zeros.
Don't you have to have one zero in order to find the other ones?

2) 1/x² - 2/x - 1 = 0 **

1 - 2x - x² / x² = 0 No, this line should just be 1 - 2x - x² = 0

I get stuck here. Any suggestions?

Thanks for the help!

x cannot be zero. Multiply each side by \(\displaystyle x^2:\)


\(\displaystyle (x^2)(1/x^2 - 2/x - 1) \ = \ (x^2)(0)\)


\(\displaystyle 1 - 2x - x^2 \ = \ 0\)

 

[megadeth95]

Thanks for the help Denis, wjm11, and lookagain