Line and Parabola Intersection NEED HELP ASAP

[mathhelp92]

Determine the equations of lines with slope 2 and that intersect the quadratic function f(x)=x(6-x) twice, once, and never

 

[matt000r000]

well, first off, f(x)=(6-x) is not a quadratic function. it must have "x^2" somewhere in it. i bet you just wrote it wrong on the website.
PS- ooops, i guess i didn't read it correctly...

for the no-intersect line, i would guess rediculously low or high, depending on which way the parabola went. for the double-intersect line, i do the same thing but in the opposite direction as the no-intersect. not the most mathematical approach, but it works. as for the single intersect, i have no clue myself. you'll have to wait for someone else to answer that.

 

[soroban]

Hello, mathhelp92!

Determine the equations of lines with slope 2 and that intersect the quadratic function:
$\displaystyle f(x)\:=\:x(6-x)$ twice, once, and never.

$\displaystyle \text{A line with slope 2 has the equation: }\:y \:=\:2x + b\text{, where }b\text{ is the }y\text{-intercept.}$

$\displaystyle \text{It intersects the parabola }y \:=\:6x-x^2\text{ when: }\:2x + b \:=\:6x-x^2$

. . $\displaystyle \text{and we have: }\:x^2 - 4x + b \:=\:0$

$\displaystyle \text{Quadratic Formula: }\;x \:=\:\frac{4 \pm\sqrt{4^2-4b}}{2} \;=\;2 \pm\sqrt{4-b}$


A quadratic equation has two roots when its discriminant is positive: .$\displaystyle 4-b \:>\>0 \quad\Rightarrow\quad b \:<\:4$

A quadratic equation has one root when its discriminant is zero: .$\displaystyle 4-b \:=\:0\quad\Rightarrow\quad b \:=\:4$

A quadratic equation has no roots when its discriminant is negative: .$\displaystyle 4-b \:<\:0 \quad\Rightarrow\quad b \:>\:4$


$\displaystyle \text{The line }y \:=\:2x + b\text{ intersects the parabola: }\:\begin{Bmatrix}\text{twice} & \text{if }b\:<\:4 \\ \text{once} & \text{if }b \:=\:4 \\ \text{never} & \text{if }b \:>\:4 \end{Bmatrix}$