Answer :
Let's examine each statement about the graph of the piecewise function [tex]\( f \)[/tex] to determine if it is true or false.
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex]:
- To find where the graph crosses the [tex]\( y \)[/tex]-axis, we need to evaluate [tex]\( f \)[/tex] at [tex]\( x = 0 \)[/tex].
- Using the middle piece of the function [tex]\( 4x - 15 \)[/tex] when [tex]\( -2 \leq x < 4 \)[/tex], substitute [tex]\( x = 0 \)[/tex] to get [tex]\( f(0) = 4(0) - 15 = -15 \)[/tex].
- Thus, the graph does cross the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex], making the statement true.
2. The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex]:
- To investigate discontinuity at [tex]\( x = -2 \)[/tex], we evaluate both the left and middle pieces of the function at [tex]\( x = -2 \)[/tex].
- For the left piece, [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex], substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = 23 \][/tex]
- For the middle piece, [tex]\( f(x) = 4x - 15 \)[/tex], substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 4(-2) - 15 = -23 \][/tex]
- Since [tex]\( 23 \neq -23 \)[/tex], the function is discontinuous at [tex]\( x = -2 \)[/tex], making the statement true.
3. The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex]:
- For [tex]\( x > 4 \)[/tex], the function is [tex]\( f(x) = 3^{x-4} \)[/tex].
- Observing the exponential function [tex]\( 3^{x-4} \)[/tex], it is clear that it increases as [tex]\( x \)[/tex] increases.
- Hence, the graph is increasing over the interval [tex]\( (4, \infty) \)[/tex], making the statement true.
4. The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex]:
- For [tex]\( x < -2 \)[/tex], the function is [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex].
- We need to check if [tex]\( f(x) \)[/tex] decreases within the interval [tex]\( (-12, -2) \)[/tex]. We evaluate [tex]\( f \)[/tex] at points [tex]\( x = -4 \)[/tex] and [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-4) = -\frac{1}{4}(-4)^2 + 6(-4) + 36 = 4 \][/tex]
[tex]\[ f(-3) = -\frac{1}{4}(-3)^2 + 6(-3) + 36 = 16.75 \][/tex]
- Since [tex]\( 4 < 16.75 \)[/tex], the graph is indeed decreasing in this interval, making the statement true.
5. The domain of the function is all real numbers:
- The function [tex]\( f \)[/tex] is defined for all real numbers [tex]\( x \)[/tex] in the given domain.
- Therefore, the statement that the domain is all real numbers is true.
Based on the analysis, we have:
\begin{tabular}{|l|c|c|}
\hline The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0,-15) \)[/tex] & true & false \\
\hline The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex] & true & false \\
\hline The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex] & true & false \\
\hline The graph is decreasing over the interval [tex]\( (-12,-2) \)[/tex] & true & false \\
\hline The domain of the function is all real numbers & true & false \\
\hline
\end{tabular}
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex]:
- To find where the graph crosses the [tex]\( y \)[/tex]-axis, we need to evaluate [tex]\( f \)[/tex] at [tex]\( x = 0 \)[/tex].
- Using the middle piece of the function [tex]\( 4x - 15 \)[/tex] when [tex]\( -2 \leq x < 4 \)[/tex], substitute [tex]\( x = 0 \)[/tex] to get [tex]\( f(0) = 4(0) - 15 = -15 \)[/tex].
- Thus, the graph does cross the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex], making the statement true.
2. The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex]:
- To investigate discontinuity at [tex]\( x = -2 \)[/tex], we evaluate both the left and middle pieces of the function at [tex]\( x = -2 \)[/tex].
- For the left piece, [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex], substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = 23 \][/tex]
- For the middle piece, [tex]\( f(x) = 4x - 15 \)[/tex], substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 4(-2) - 15 = -23 \][/tex]
- Since [tex]\( 23 \neq -23 \)[/tex], the function is discontinuous at [tex]\( x = -2 \)[/tex], making the statement true.
3. The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex]:
- For [tex]\( x > 4 \)[/tex], the function is [tex]\( f(x) = 3^{x-4} \)[/tex].
- Observing the exponential function [tex]\( 3^{x-4} \)[/tex], it is clear that it increases as [tex]\( x \)[/tex] increases.
- Hence, the graph is increasing over the interval [tex]\( (4, \infty) \)[/tex], making the statement true.
4. The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex]:
- For [tex]\( x < -2 \)[/tex], the function is [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex].
- We need to check if [tex]\( f(x) \)[/tex] decreases within the interval [tex]\( (-12, -2) \)[/tex]. We evaluate [tex]\( f \)[/tex] at points [tex]\( x = -4 \)[/tex] and [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-4) = -\frac{1}{4}(-4)^2 + 6(-4) + 36 = 4 \][/tex]
[tex]\[ f(-3) = -\frac{1}{4}(-3)^2 + 6(-3) + 36 = 16.75 \][/tex]
- Since [tex]\( 4 < 16.75 \)[/tex], the graph is indeed decreasing in this interval, making the statement true.
5. The domain of the function is all real numbers:
- The function [tex]\( f \)[/tex] is defined for all real numbers [tex]\( x \)[/tex] in the given domain.
- Therefore, the statement that the domain is all real numbers is true.
Based on the analysis, we have:
\begin{tabular}{|l|c|c|}
\hline The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0,-15) \)[/tex] & true & false \\
\hline The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex] & true & false \\
\hline The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex] & true & false \\
\hline The graph is decreasing over the interval [tex]\( (-12,-2) \)[/tex] & true & false \\
\hline The domain of the function is all real numbers & true & false \\
\hline
\end{tabular}
