1. ## Finding the interval of expression having two quadratic equations.

What will be the values of 'm' so that the range of the equation
$\displaystyle y= \frac{mx^2+3x-4}{-4x^2+3x+m}$

will be all real values( i.e. $\displaystyle y\epsilon [-\infty,\infty]$ ).

given:x can take all real values.
any help or hint will be appreciated.

2. ## Re: Finding the interval of expression having two quadratic equations.

You might try the Quadratic Formula. Parts of it may tell you where the denominator has no Real Solution.

3. ## Re: Finding the interval of expression having two quadratic equations.

But even if the denominator has no real solution then also any real value of x will give some value of the denominator and hence will give some value of the whole equation so I think that we don't have to find the solutions or something related to solutions

4. ## Re: Finding the interval of expression having two quadratic equations.

Whoops! I read "Domain", somehow, where it says "Range".

Given that the Domain IS the Real Numbers, 3^2 + 4*4*m < 0 or 16m < -9 or m < -9/16

In the basic rule list of rational functions, the ONLY way to get ALL Real Numbers in the Range is to avoid Horizontal Asymptotes. With equal or lesser degree in the numerator than in the denominator, how can we do that? We can't. Well, we can also get two vertical asymptotes and pick up the last value in the middle section. Too bad we don't get any vertical asymptotes on this assignment.

Where does that leave us?

5. ## Re: Finding the interval of expression having two quadratic equations.

but the answer to this is
$\displaystyle m\epsilon [1,7]$
so i am confused

6. ## Re: Finding the interval of expression having two quadratic equations.

No, that's not it. It is as I stated. Are you SURE is says "Given: x can take on all Real values." If it says that, then [1,7] is most defintely NOT a solution set to this question. [1,7] does do what I said earlier, "...we can also get two vertical asumptotes and pick up the last value in the middle section." However, like I also said earlier, "Too bad we don't get any vertical asymptotes on this assignment." Vertical asymptotes are inconsistent with "Given: x can take on all Real values." I should also point out that x = 7 and x = 1 are not solutions even if we get vertical asymptotes.

You do not need to be confused. You just have to get used to arguing with the answer key when it is wrong.

ok thank you