Question

Answer the questions about each function. $$f(x)=\frac{x^{2}+2}{x+4}$$ (a) Is the point \(\left(1, \frac{3}{5}\right)\) on the graph of \(f ?\) (b) If \(x=0,\) what is \(f(x) ?\) What point is on the graph of \(f ?\) (c) If \(f(x)=\frac{1}{2},\) what is \(x ?\) What point \((\mathrm{s})\) are on the graph of \(f ?\) (d) What is the domain of \(f ?\) (e) List the \(x\) -intercepts, if any, of the graph of \(f\). (f) List the \(y\) -intercept, if there is one, of the graph of \(f\).

Short answer

The point (1, 3/5) is on the graph. f(0) = 1/2, point (0, 1/2) is on the graph. Solving f(x) = 1/2 gives x = 0 or x = 1/2, points (0, 1/2) and (1/2, 1/2) are on the graph. The domain is all real numbers except x = -4. There are no x-intercepts. The y-intercept is (0, 1/2).

Step-by-step solution

- Check if point (1, 3/5) is on the graph of f

Substitute x = 1 into the function and check if the output is equal to 3/5. Calculate: \[f(1) = \frac{1^2 + 2}{1 + 4} = \frac{3}{5}\] Since \(f(1) = \frac{3}{5}\), the point \( (1, \frac{3}{5}) \) is on the graph of \(f\).

- Calculate f(0) and identify the point on the graph

Substitute x = 0 into the function: \[f(0) = \frac{0^2 + 2}{0 + 4} = \frac{2}{4} = \frac{1}{2}\] The point corresponding to x = 0 on the graph of f is (0, 1/2).

- Solve for x when f(x) = 1/2

Set the function equal to 1/2 and solve for x: \[ \frac{x^2 + 2}{x + 4} = \frac{1}{2} \] Cross-multiplying gives: \[ 2(x^2 + 2) = x + 4 \] \[ 2x^2 + 4 = x + 4 \] \[ 2x^2 - x = 0 \] Factor out x: \[ x(2x - 1) = 0\] So, x = 0 or x = 1/2. The points are (0, 1/2) and (1/2, 1/2).

- Determine the domain of f

Find values of x that make the denominator zero. The denominator is x + 4, so: \[ x + 4 = 0 \] \[ x = -4 \] The domain of f is all real numbers except x = -4.

- Find the x-intercepts

Set f(x) = 0 and solve for x: \[ \frac{x^2 + 2}{x + 4} = 0 \] The numerator must equal zero: \[ x^2 + 2 = 0 \] This has no real solutions, so there are no x-intercepts.

- Find the y-intercept

Substitute x = 0 to find the y-intercept: \[f(0) = \frac{1}{2}\] The y-intercept is the point (0, 1/2).

Key concepts

Function Graph Analysis

When analyzing a rational function graph, it's essential to look at several key features such as points on the graph, intercepts, and any asymptotic behavior. For the function given by \( f(x) = \frac{x^2 + 2}{x + 4} \), each of these can tell us important details about its behavior.

First, we check if specific points lie on the graph. For example, substitute \( x = 1 \) and see if the resulting \( f(x) \) equals any given y-coordinate. Here, \( f(1) = \frac{1^2 + 2}{1 + 4} = \frac{3}{5} \). Thus, the point \( (1, \frac{3}{5}) \) is on the graph.

It’s also helpful to examine what happens for specific values of \( x \). For instance, \( f(0) = \frac{0^2 + 2}{0 + 4} = \frac{1}{2} \), showing that the point \( (0, \frac{1}{2}) \) is on the graph. This can help understand intercepts and solve for unknowns.

Domain and Range

The domain of a rational function consists of all values of \( x \) that do not make the denominator zero. For \( f(x) = \frac{x^2 + 2}{x+4} \), the denominator becomes zero at \( x = -4 \). Thus, the domain of \( f \) is all real numbers except \( x = -4 \). We write it as \( x e -4 \).

The range of the function is given by the set of all possible outputs, \( f(x) \). To find the range, it's usually beneficial to consider the behavior of the function as \( x \) approaches various critical points (like where the denominator is zero) and as \( x \) approaches infinity. For many rational functions, some values may not be attained, which can be verified through closer analysis or graphical methods.

Intercepts

Intercepts are points where the graph crosses the axes. For the given function, we find these as follows:

x-intercepts: To find x-intercepts, set \( f(x) = 0 \). This implies the numerator must be zero, because zero divided by any non-zero number is zero. Thus, solve \( x^2 + 2 = 0 \). This equation has no real solutions (since squares of real numbers are non-negative and \( x^2 \) cannot be -2), so there are no x-intercepts.

y-intercept: The y-intercept is where the graph crosses the y-axis, which happens when \( x = 0 \). Here, \( f(0) = \frac{2}{4} = \frac{1}{2} \). Hence, the y-intercept is \( (0, \frac{1}{2}) \). Finding these intercepts is crucial in sketching the basic shape of the graph and understanding the behavior of the function at critical points.