The denominator cannot be zero.  Set the denominator not equal to zero to determine which values of \(\displaystyle x\) will not be part of the domain.  

\(\displaystyle (5-x)(x-3)\neq0\)

Split the terms, and solve for \(\displaystyle x\).

\(\displaystyle x-3\neq0\)

Add three on both sides.

\(\displaystyle x\neq3\)

Set the other binomial equal to zero.

\(\displaystyle 5-x\neq0\)

Add \(\displaystyle x\) on both sides.

\(\displaystyle x\neq5\)

The domain cannot exist at \(\displaystyle x=3, 5\) and asymptotes will exist at those two values.

The answer in interval notation is:  

\(\displaystyle (-\infty, 3)\bigcup(3,5)\bigcup(5,\infty)\)

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:  \(\displaystyle \textup{All real numbers.}\)

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

Recall that the range and domain of the parent function \(\displaystyle y=\sqrt x\) only exists when \(\displaystyle x\geq0, y\geq0\).

Applying the scale factor and the horizontal shifts will not affect the range of the curve.  Only the constant \(\displaystyle -30\) after the \(\displaystyle 30\sqrt{x-30}\) term will affect the vertical shift of the graph.

This constant will shift the graph down 30 units.

The range is:  \(\displaystyle y\geq-30\).

The answer in interval notation is:  \(\displaystyle [-30,\infty)\)

The parent function \(\displaystyle y=ln(x)\) has a domain of \(\displaystyle x>0\) since there is an asymptote at \(\displaystyle x=0\).

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

The \(\displaystyle (x+2)\) term will shift the asymptote and the line of natural log left two units.

This means that the domain of \(\displaystyle y=ln(x+2)\) will be \(\displaystyle x>-2\).

The answer is:  \(\displaystyle (-2,\infty)\)

Distribute the negative two through the binomial.

\(\displaystyle y=-2(x^2-4) = -2(x^2)+(-2)(-4) = -2x^2+8\)

The parabolic equation in standard form is:   \(\displaystyle y=ax^2+bx+c\)

\(\displaystyle y= -2x^2+8\)

Notice that since we have a negative coefficient, the parabola will open downward.  The vertex is centered at \(\displaystyle x=0\).  The y-intercept is 8, which means this is a maximum y-value of the parabola.

The answer is:  \(\displaystyle (-\infty, 8]\)

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.

\(\displaystyle 5-3x\geq0\)

Solve for x.  Add \(\displaystyle 3x\) on both sides.

\(\displaystyle 5\geq3x\)

Divide by three on both sides.

\(\displaystyle x\leq\frac{5}{3}\)

The answer is:  \(\displaystyle x\leq\frac{5}{3}\)

Range is the \(\displaystyle y\) value generated from a real \(\displaystyle x\) value. We know square roots have to generate all values greater than or equal to zero. Therefore the answer is \(\displaystyle 0\leq y\)regardless of the function inside the radical unless there is a fraction present. 

Range is the \(\displaystyle y\) value generated from a real \(\displaystyle x\) 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. 

\(\displaystyle f(x)=\frac{6}{x^2-25}\) 

We only need to look at \(\displaystyle x^2-25\).

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

\(\displaystyle x^2-25=0\) Add \(\displaystyle 25\) on both sides.

\(\displaystyle x^2=25\) Take the square root of both sides. Remember it can also be negative.

\(\displaystyle x=\pm5\) Let's check values of \(\displaystyle \pm4.99, \pm5.01\). Let's analyze the \(\displaystyle \pm4.99\) values. We do this to find the ranges.

If \(\displaystyle x=\pm4.99\), we get a small denominator value that is negative. This means the answer is negative infinity. If \(\displaystyle x=\pm5.01\), we get a small positive denominator value. This means the answer is positive infinity. Let's see when \(\displaystyle x\) is \(\displaystyle \pm \infty\).

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

The parent function of \(\displaystyle x^4\) looks similar to the \(\displaystyle x^2\) graph and will open upwards.

The negative coefficient in front of the \(\displaystyle x^4\) term indicates that the graph will open downward, which means that the lowest range value is negative infinity.

The y-intercept is 9, an will be the highest point on this graph.

The range is:  \(\displaystyle (-\infty,9]\)

Notice that this is a parabolic function that will open downward.  The domain refers to all possible x-values on the graph.  

The parent function \(\displaystyle y=x^2\) has a domain of all real numbers and a range from \(\displaystyle [0,\infty)\).  The transformations of \(\displaystyle y=\frac{1}{5}(9-x^2)+6\) will not affect the domain, but the range of the graph since the y-values of the graph are affected.

There are no values of the x-variable that will make this function undefined, which means all real numbers can exist.

The answer is:  \(\displaystyle (-\infty,\infty)\)