f(x)=(5x^2-7x+2)/(4x^2-2)

f(x)--> ? as x--> -infinity

f(x)--> ? as x--> infinty

can someone help me out here, i seem to be coming up with the wrong answers..thanks

2. $\frac{5x^2-7x+2}{4x^2-2}$

We can try to make the denominator 0.

$4x^2-2=0$

$x^2 = \frac{1}{2}$

$x = \{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \}$

Now, plug those values in the function and check which one makes it $+\infty$ or $- \infty$

3. Originally Posted by mathlete
f(x)=(5x^2-7x+2)/(4x^2-2)

f(x)--> ? as x--> -infinity

f(x)--> ? as x--> infinty

can someone help me out here, i seem to be coming up with the wrong answers..thanks
$f(x) = \frac{5x^2 - 7x + 2}{4x^2 - 2}$.

Divide numerator and denominator by $x^2$:

$f(x) = \frac{5 - \frac{7}{x} + \frac{2}{x^2}}{4 - \frac{2}{x^2}}$.

Therefore:

As $x \rightarrow \pm \infty$, $f(x) \rightarrow \frac{5 - 0 + 0}{4 - 0} = \frac{5}{4}$.

4. Originally Posted by wingless
$\frac{5x^2-7x+2}{4x^2-2}$

We can try to make the denominator 0.

$4x^2-2=0$

$x^2 = \frac{1}{2}$

$x = \{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \}$

Now, plug those values in the function and check which one makes it $+\infty$ or $- \infty$
You are making $\mathrm{f}(x)\to\pm\infty$. The question is about $x\to\pm\infty$.

5. Haha sorry I just misread it