Answer :
Let's analyze each statement about the function [tex]\( f \)[/tex] given by:
[tex]\[ f(x)=\left\{ \begin{array}{ll} -\frac{1}{4} x^2+6 x+36, & x < -2 \\ 4 x-15, & -2 \leq x < 4 \\ 3^{x-4}, & x > 4 \\ \end{array} \right. \][/tex]
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
To determine if the graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex], we need to evaluate [tex]\( f(0) \)[/tex].
For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 4 \cdot 0 - 15 = -15 \][/tex]
Since [tex]\( f(0) = -15 \)[/tex], the graph indeed crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
2. The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex].
To check for a point of discontinuity at [tex]\( x = -2 \)[/tex], we need to evaluate the left-hand limit and the right-hand limit at [tex]\( x = -2 \)[/tex].
Left-hand limit as [tex]\( x \to -2^- \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} f(x) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = -1 - 12 + 36 = 23 \][/tex]
Right-hand limit as [tex]\( x \to -2^+ \)[/tex]:
[tex]\[ \lim_{{x \to -2^+}} f(x) = 4(-2) - 15 = -8 - 15 = -23 \][/tex]
Since the left-hand limit (23) is not equal to the right-hand limit (-23), there is a point of discontinuity at [tex]\( x = -2 \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
3. The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
For [tex]\( x > 4 \)[/tex], the function is given by [tex]\( f(x) = 3^{x-4} \)[/tex]. Exponential functions of the form [tex]\( 3^{x-4} \)[/tex] are strictly increasing.
Therefore, the graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
4. The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex].
On the interval [tex]\( (-12, -2) \)[/tex], the function is given by [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex].
To determine if the function is decreasing, we need to check the derivative of this part and ensure it's negative over [tex]\( (-12, -2) \)[/tex].
Evaluating the derivative:
[tex]\[ f'(x) = -\frac{1}{2} x + 6 \][/tex]
At [tex]\( x = -2 \)[/tex]:
[tex]\[ f'(-2) = -\frac{1}{2}(-2) + 6 = 1 + 6 = 7 \][/tex]
Since [tex]\( f'(x) > 0 \)[/tex] in the interval [tex]\( (-12, -2) \)[/tex], the function is increasing, not decreasing.
[tex]\[ \text{Answer: false} \][/tex]
5. The domain of the function is all real numbers.
The function [tex]\( f \)[/tex] is defined for all [tex]\( x \)[/tex] in the real numbers, as each piece of the function covers all real intervals without any restrictions or undefined values.
[tex]\[ \text{Answer: true} \][/tex]
Summarizing the answers:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{The graph crosses the } y \text{-axis at } (0,-15). & \text{true} & \text{false} \\ \hline \text{The graph has a point of discontinuity at } x =-2. & \text{true} & \text{false} \\ \hline \text{The graph is increasing over the interval } (4, \infty). & \text{true} & \text{false} \\ \hline \text{The graph is decreasing over the interval } (-12,-2). & \text{true} & \text{false} \\ \hline \text{The domain of the function is all real numbers}. & \text{true} & \text{false} \\ \hline \end{array} \][/tex]
[tex]\[ f(x)=\left\{ \begin{array}{ll} -\frac{1}{4} x^2+6 x+36, & x < -2 \\ 4 x-15, & -2 \leq x < 4 \\ 3^{x-4}, & x > 4 \\ \end{array} \right. \][/tex]
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
To determine if the graph crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex], we need to evaluate [tex]\( f(0) \)[/tex].
For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 4 \cdot 0 - 15 = -15 \][/tex]
Since [tex]\( f(0) = -15 \)[/tex], the graph indeed crosses the [tex]\( y \)[/tex]-axis at [tex]\( (0, -15) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
2. The graph has a point of discontinuity at [tex]\( x = -2 \)[/tex].
To check for a point of discontinuity at [tex]\( x = -2 \)[/tex], we need to evaluate the left-hand limit and the right-hand limit at [tex]\( x = -2 \)[/tex].
Left-hand limit as [tex]\( x \to -2^- \)[/tex]:
[tex]\[ \lim_{{x \to -2^-}} f(x) = -\frac{1}{4}(-2)^2 + 6(-2) + 36 = -1 - 12 + 36 = 23 \][/tex]
Right-hand limit as [tex]\( x \to -2^+ \)[/tex]:
[tex]\[ \lim_{{x \to -2^+}} f(x) = 4(-2) - 15 = -8 - 15 = -23 \][/tex]
Since the left-hand limit (23) is not equal to the right-hand limit (-23), there is a point of discontinuity at [tex]\( x = -2 \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
3. The graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
For [tex]\( x > 4 \)[/tex], the function is given by [tex]\( f(x) = 3^{x-4} \)[/tex]. Exponential functions of the form [tex]\( 3^{x-4} \)[/tex] are strictly increasing.
Therefore, the graph is increasing over the interval [tex]\( (4, \infty) \)[/tex].
[tex]\[ \text{Answer: true} \][/tex]
4. The graph is decreasing over the interval [tex]\( (-12, -2) \)[/tex].
On the interval [tex]\( (-12, -2) \)[/tex], the function is given by [tex]\( f(x) = -\frac{1}{4} x^2 + 6 x + 36 \)[/tex].
To determine if the function is decreasing, we need to check the derivative of this part and ensure it's negative over [tex]\( (-12, -2) \)[/tex].
Evaluating the derivative:
[tex]\[ f'(x) = -\frac{1}{2} x + 6 \][/tex]
At [tex]\( x = -2 \)[/tex]:
[tex]\[ f'(-2) = -\frac{1}{2}(-2) + 6 = 1 + 6 = 7 \][/tex]
Since [tex]\( f'(x) > 0 \)[/tex] in the interval [tex]\( (-12, -2) \)[/tex], the function is increasing, not decreasing.
[tex]\[ \text{Answer: false} \][/tex]
5. The domain of the function is all real numbers.
The function [tex]\( f \)[/tex] is defined for all [tex]\( x \)[/tex] in the real numbers, as each piece of the function covers all real intervals without any restrictions or undefined values.
[tex]\[ \text{Answer: true} \][/tex]
Summarizing the answers:
[tex]\[ \begin{array}{|l|l|l|} \hline \text{The graph crosses the } y \text{-axis at } (0,-15). & \text{true} & \text{false} \\ \hline \text{The graph has a point of discontinuity at } x =-2. & \text{true} & \text{false} \\ \hline \text{The graph is increasing over the interval } (4, \infty). & \text{true} & \text{false} \\ \hline \text{The graph is decreasing over the interval } (-12,-2). & \text{true} & \text{false} \\ \hline \text{The domain of the function is all real numbers}. & \text{true} & \text{false} \\ \hline \end{array} \][/tex]
