Expand the quadratic.  

Use the FOIL method to expand the binomials.

The equation becomes:

Now that we have the equation in  format, find the vertex.  This will determine the minimum of the parabola.

The formula is:

Substitute the values.

To find the y-value, substitute the x-value back to the original equation.

The minimum is:  

Because the value of  is positive, the parabola will open up.

The range is:  

The range is the existing y-values that contains the function.

Notice that this is a parabola that opens downward, and the y-intercept is four.

This means that the highest y-value on this graph is four.  The y-values will approach negative infinity as the domain, or x-values, approaches to positive and negative infinity.

The answer is:  

The denominator cannot be zero.  Set the denominator not equal to zero to determine which values of  will not be part of the domain.  

Split the terms, and solve for .

Add three on both sides.

Set the other binomial equal to zero.

Add  on both sides.

The domain cannot exist at  and asymptotes will exist at those two values.

The answer in interval notation is:  

The given equation does not have any x-values that are bounded by any restrictions.  There is no denominator in the equation, and the x-value may contain all real numbers for its input.

The answer is:  

The range consists of all valid y-values that a function can possibly have.

Recall that the range and domain of the parent function  only exists when .

Applying the scale factor and the horizontal shifts will not affect the range of the curve.  Only the constant  after the  term will affect the vertical shift of the graph.

This constant will shift the graph down 30 units.

The range is:  .

The answer in interval notation is:  

The parent function  has a domain of  since there is an asymptote at .

There cannot be a natural log of zero or a negative value.

The  term will shift the asymptote and the line of natural log left two units.

This means that the domain of  will be .

The answer is:  

Distribute the negative two through the binomial.

The parabolic equation in standard form is:   

Notice that since we have a negative coefficient, the parabola will open downward.  The vertex is centered at .  The y-intercept is 8, which means this is a maximum y-value of the parabola.

The answer is:  

Notice that the quantity of what's inside square root cannot be a negative number.

Set the inner quantity such that it must be greater than or equal to zero.

Solve for x.  Add  on both sides.

Divide by three on both sides.

The answer is:  

Range is the  value generated from a real  value. We know square roots have to generate all values greater than or equal to zero. Therefore the answer is regardless of the function inside the radical unless there is a fraction present. 

Range is the  value generated from a real  value. Because this function is a fractional expression, we need to check the denominator. Remember, the denominator must not be zero. This would make the function undefined. 

We only need to look at .

Since we know it can't equal zero, we set that expression to zero to determine the -value that makes this function undefined.

 Add  on both sides.

 Take the square root of both sides. Remember it can also be negative.

 Let's check values of . Let's analyze the  values. We do this to find the ranges.

If , we get a small denominator value that is negative. This means the answer is negative infinity. If , we get a small positive denominator value. This means the answer is positive infinity. Let's see when  is .

We then get values extremely small and approaching zero but NEVER BEING ZERO. Therefore the answer is all real numbers except zero.