Determine the equations of lines with slope 2 and that intersect the quadratic function f(x)=x(6-x)?

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- Mar 18th 2009, 02:50 PMmathhelp92parabolas and line intersection
Determine the equations of lines with slope 2 and that intersect the quadratic function f(x)=x(6-x)?

- Mar 18th 2009, 03:18 PMskeeter
let the lines be $\displaystyle y = 2x + k$, where $\displaystyle k$ is the y-intercept

$\displaystyle 2x+k = x(6-x)

$

$\displaystyle 2x+k = 6x - x^2$

$\displaystyle x^2 - 4x + k = 0$

to have real solutions, $\displaystyle b^2 - 4ac \geq 0$

$\displaystyle 16 - 4k \geq 0$

$\displaystyle k \leq 4$ - Mar 18th 2009, 03:28 PMmathhelp92
- Mar 18th 2009, 03:32 PMProve It
- Mar 18th 2009, 03:45 PMmathhelp92

So if k > 4, so then in the equation y=mx+b it will be y=2x+4..I THINK

so thats 2 roots

so how do we solve for only one root or no roots

because if we make both equations equal to each other, we will get 4 every time :s this is where i am confused

WHAT WE ARE TRYING TO SOLVE FOR IS THE VALUE OF B IN y=mx+b for 3 equations soo 3 different values of b - Mar 18th 2009, 04:31 PMProve It
I've already told you the answer.

If there aren't any intersection points then the value of $\displaystyle b$ is ANYTHING LESS THAN 4. So use a letter like $\displaystyle k$ and say that $\displaystyle k <4$.

If there is only one intersection point then the value of $\displaystyle b$ IS 4.

If there are two intersection points then the value of $\displaystyle b$ is ANYTHING GREATER THAN 4. So use a letter like $\displaystyle k$ and say that $\displaystyle k > 4$.