please help!!

find the set for "m" so that the graphs of y=mx and y=(x^2)+m intersect in a single point.

- Sep 5th 2007, 05:20 PMginax7Find the set for "m" so that the graphs of y=mx and y=x^2+m intersect a single poi
please help!!

find the set for "m" so that the graphs of y=mx and y=(x^2)+m intersect in a single point.

- Sep 5th 2007, 05:27 PMKrizalid
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

$\displaystyle m^2-4m=0\implies m=0,4$ - Sep 5th 2007, 05:28 PMtopsquark
ginax7,

*please*consider using a different colored font. You know, black really is an acceptable color for typing. Really! :D

Quote:

Find 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 - Sep 5th 2007, 05:32 PMginax7thanks
thanks guys! sorry bout the pinkness! It's the first day of school and I havent even learned about discriminants I dunno why hes giving us problems like these :(

- Sep 5th 2007, 05:38 PMKrizalid
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 :)