• Mar 23rd 2009, 07:26 AM
princess_21
what if the radical is in the denominator?
$\displaystyle y=\frac{2}{x^2+8x+10}$

the domain is $\displaystyle \{x|x^2+8x+10\ge 0\}$

so the domain is $\displaystyle x^2+8x+10\ge0$

$\displaystyle \frac{4ac-b^2}{4a}$

$\displaystyle \frac{40-64}{4}$

$\displaystyle -4$

$\displaystyle x\ge-4$??

range: ??

is this correct?
• Mar 23rd 2009, 07:58 AM
u2_wa
First check where $\displaystyle x^2+8x+10=0$
Solution is $\displaystyle x= -6.5$ and $\displaystyle -1.6$

This means as domain is $\displaystyle x\ge0$ so the equation $\displaystyle x^2+8x+10$ is always positive.

This means if $\displaystyle x=0$,y= $\displaystyle \frac{2}{0^2+8(0)+10}$ $\displaystyle = \frac{1}{5}$

As $\displaystyle x^2+8x+10$ always increases when increases, So the value of this function $\displaystyle \frac{2}{x^2+8x+10}$ decreases.
Hence max. value this function can have is $\displaystyle \frac{1}{5}$ and the graph of this function tends to zero, when x increases but can never be equal to zero.

Range :$\displaystyle \frac{1}{5}\ge y>0$
• Mar 23rd 2009, 08:07 AM
princess_21
• Mar 23rd 2009, 10:05 AM
stapel
Quote:

Originally Posted by princess_21
how did you get x = -6.5 and x = -1.6?

The poster solved the quadratic equation.

You can, for instance, use the Quadratic Formula for this.... (Wink)
• Mar 24th 2009, 01:34 AM
u2_wa
Quote:

Originally Posted by princess_21

I have used this formula to solve $\displaystyle x^2+8x+10=0$ :

$\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where standard equation is this one: $\displaystyle ax^2+bx+c$
a=1, b=8, c=10

Plug in the values to solve the equation.