Determine the equations of lines with slope 2 and that intersect the quadratic function f(x)=x(6-x)?
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, $\displaystyle b^2 - 4ac \geq 0$
$\displaystyle 16 - 4k \geq 0$
$\displaystyle k \leq 4$
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
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 $\displaystyle k > 4$.