please help!!
find the set for "m" so that the graphs of y=mx and y=(x^2)+m intersect in a single point.
First, we gotta set $\displaystyle f(x)=mx$ and $\displaystyle g(x)=x^2+m$, now set $\displaystyle f(x)=g(x)$
$\displaystyle x^2+m-mx=0\implies x^2-mx+m=0$
The discriminant of this must be zero (such that $\displaystyle f(x)$ and $\displaystyle g(x)$ intersect in a single point), so
$\displaystyle m^2-4m=0\implies m=0,4$
ginax7, please consider using a different colored font. You know, black really is an acceptable color for typing. Really!
LetFind the set for "m" so that the graphs of y = mx and y = (x^2)+m intersect in a single point.
$\displaystyle mx = x^2 + m$
$\displaystyle x^2 - mx + m = 0$
This must have only one solution, so the discriminant must be 0:
$\displaystyle (-m)^2 + 4m = 0$
$\displaystyle m^2 - 4m = 0$
$\displaystyle m(m - 4) = 0$
So either m = 0 or m = 4.
-Dan
Suppose you have the following quadratic equation $\displaystyle ax^2+bx+c=0,\,\forall a,b,c\in\mathbb R,\,a\ne0$
The following formula solves this equation
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
The expression $\displaystyle \Delta=b^2-4ac$ it's called discriminant, which says who rules here!
If $\displaystyle \Delta>0$, then the equation has two real and different roots.
If $\displaystyle \Delta=0$, the the equation has two real and equal roots
If $\displaystyle \Delta<0$, the equation has two complex roots.
Hope it helps