# parabolas and line intersection

• Mar 18th 2009, 02:50 PM
mathhelp92
parabolas 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 PM
skeeter
Quote:

Originally Posted by mathhelp92
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, $\displaystyle b^2 - 4ac \geq 0$

$\displaystyle 16 - 4k \geq 0$

$\displaystyle k \leq 4$
• Mar 18th 2009, 03:28 PM
mathhelp92
Quote:

Originally Posted by skeeter
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, $\displaystyle b^2 - 4ac \geq 0$

$\displaystyle 16 - 4k \geq 0$

$\displaystyle k \leq 4$

THANK YOU...but this way I will get the roots of it
i need the equations for the line that have a slope of 2 and they pass through the parabola once, never n twice
• Mar 18th 2009, 03:32 PM
Prove It
Quote:

Originally Posted by mathhelp92
THANK YOU...but this way I will get the roots of it
i need the equations for the line that have a slope of 2 and they pass through the parabola once, never n twice

To have 0 real solutions, the discriminant is negative. Solve for k.

To have 1 real solution, the discriminant is 0. Solve for k.

To have 2 real solutions, the discriminant is positive. Solve for k.
• Mar 18th 2009, 03:45 PM
mathhelp92
Quote:

Originally Posted by Prove It
To have 0 real solutions, the discriminant is negative. Solve for k.

To have 1 real solution, the discriminant is 0. Solve for k.

To have 2 real solutions, the discriminant is positive. Solve for k.

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 PM
Prove It
Quote:

Originally Posted by mathhelp92
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

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 $\displaystyle k > 4$.