# What is the domain and range of 5/(4(x+2)^2) -1?

domain: $x > - 2$ or $x < - 2$ range: f(x)>-1

#### Explanation:

we should take a look at the denominator and notice that it resets if x=-2.

the domain: we'll compare the denominator to 0 to get $4 {\left(x + 2\right)}^{2} = 0 :$ which gives us as written, x=-2. so the domain is $x > - 2$ or $x < - 2$

for the range: we'll find the inverse function and compare it to 0. the original function: $y = \frac{5}{4 {\left(x + 2\right)}^{2}} - 1$ the inverse(by replacing x and y): $x = \frac{5}{4 {\left(y + 2\right)}^{2}} - 1$, and we'll solve it for y, to obtain: $x \setminus : 4 {\left(y + 2\right)}^{2} = \setminus \frac{5}{4 {\left(y + 2\right)}^{2}} \setminus : 4 {\left(y + 2\right)}^{2} - 1 \setminus \cdot \setminus : 4 {\left(y + 2\right)}^{2}$.

now we get: $4 x {\left(y + 2\right)}^{2} = 5 - 4 {\left(y + 2\right)}^{2}$ which becomes $4 {\left(y + 2\right)}^{2} \left(x + 1\right) = 5$ which becomes ${\left(y + 2\right)}^{2} = \setminus \frac{5}{4 \left(x + 1\right)}$

we'll find the 2 roots:

for $y + 2 = \setminus \sqrt{\setminus \frac{5}{4 \left(x + 1\right)}}$ we get the domain f(x)>-1
and for $y + 2 = - \setminus \sqrt{\setminus \frac{5}{4 \left(x + 1\right)}}$ we get the domain f(x)>-1

so the range is f(x)>-1