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### 2.1.1: Write down the equation of the asymptote of [tex]\( g \)[/tex].
The equation of the asymptote is a horizontal line that the graph approaches but never reaches. For the function [tex]\( g(x) = 2\left(\frac{1}{2}\right)^{x+1} - 4 \)[/tex], the horizontal asymptote is [tex]\( y = -4 \)[/tex].
### Answer:
[tex]\[ y = -4 \][/tex]
### 2.1.2: Write down the coordinates of D.
The turning point (vertex) of the quadratic function [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex] can be found by observing the standard form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. Here, the vertex is at [tex]\((3, 25)\)[/tex].
### Answer:
[tex]\[ D = (3, 25) \][/tex]
### 2.1.3: Write down the range of [tex]\( f \)[/tex].
The function [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex] is a downward-opening parabola with a maximum value at the vertex [tex]\((3, 25)\)[/tex]. Therefore, the range of [tex]\( f \)[/tex] is [tex]\((-\infty, 25]\)[/tex].
### Answer:
[tex]\[ \text{Range of } f: (-\infty, 25] \][/tex]
### 2.1.4: Calculate the length of [tex]\( EB \)[/tex].
[tex]\( E \)[/tex] is the point where [tex]\( f \)[/tex] cuts the y-axis. Evaluating [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = -(0-3)^2 + 25 = -9 + 25 = 16 \][/tex]
So, [tex]\( E = (0, 16) \)[/tex].
[tex]\( B \)[/tex] is the point where [tex]\( g \)[/tex] cuts the y-axis. Evaluating [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = 2\left(\frac{1}{2}\right)^{0+1} - 4 = 2\left(\frac{1}{2}\right) - 4 = 1 - 4 = -3 \][/tex]
So, [tex]\( B = (0, -3) \)[/tex].
The length of [tex]\( EB \)[/tex] is the vertical distance between the y-coordinates of [tex]\( E \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ \text{Length of } EB = |16 - (-3)| = |16 + 3| = 19 \][/tex]
### Answer:
[tex]\[ \text{Length of } EB = 19 \][/tex]
### 2.1.5: Determine the values of [tex]\( x \)[/tex] for which [tex]\( f \)[/tex] is decreasing.
For the quadratic function [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex], it decreases when [tex]\( x \)[/tex] is to the right of the vertex. Since the vertex is at [tex]\( x = 3 \)[/tex], the function is decreasing for [tex]\( x > 3 \)[/tex].
### Answer:
[tex]\[ x \in (3, \infty) \][/tex]
### 2.1.6: Calculate the average gradient between points A and B.
Point [tex]\( A \)[/tex] is where [tex]\( f \)[/tex] cuts the x-axis. Solving [tex]\( f(x) = 0 \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ 0 = -(x-3)^2 + 25 \][/tex]
[tex]\[ (x-3)^2 = 25 \][/tex]
[tex]\[ x - 3= \pm 5 \][/tex]
Thus, the roots are [tex]\( x = 3 + 5 = 8 \)[/tex] and [tex]\( x = 3 - 5 = -2 \)[/tex]. If considering [tex]\( A \)[/tex] to be [tex]\( (8, 0) \)[/tex].
From part 2.1.4, [tex]\( B = (0, -3) \)[/tex].
The average gradient between [tex]\( A = (8, 0) \)[/tex] and [tex]\( B = (0, -3) \)[/tex]:
[tex]\[ \text{Average Gradient} = \frac{B_y - A_y}{B_x - A_x} = \frac{-3 - 0}{0 - 8} = \frac{-3}{-8} = 0.375 \][/tex]
### Answer:
[tex]\[ \text{Average Gradient} = 0.375 \][/tex]
### 2.1.7: If [tex]\( p(x) = f(x) + 2 \)[/tex], write down the coordinates of the turning point [tex]\( p \)[/tex].
The function [tex]\( p(x) = f(x) + 2 \)[/tex] is obtained by shifting [tex]\( f(x) \)[/tex] upward by 2 units. Therefore, the turning point [tex]\( D \)[/tex] of [tex]\( f \)[/tex] shifts to [tex]\( (3, 25 + 2) = (3, 27) \)[/tex].
### Answer:
[tex]\[ D_p = (3, 27) \][/tex]
### 2.1.8: Determine the value of [tex]\( k \)[/tex] for which the straight line [tex]\( y = 2x + k \)[/tex] will be a tangent to [tex]\( f \)[/tex].
To find [tex]\( k \)[/tex], we need the line [tex]\( y = 2x + k \)[/tex] to be tangent to [tex]\( f \)[/tex] at a single point. This happens where the slope of [tex]\( f \)[/tex] matches the slope of the line.
Given [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex], the derivative [tex]\( f'(x) = -2(x-3) \)[/tex]. Equating this to the slope of the line (2):
[tex]\[ -2(x-3) = 2 \][/tex]
[tex]\[ -2x + 6 = 2 \][/tex]
[tex]\[ -2x = -4 \][/tex]
[tex]\[ x = 2 \][/tex]
Now, find [tex]\( f(2) \)[/tex]:
[tex]\[ f(2) = -(2-3)^2 + 25 = -1 + 25 = 24 \][/tex]
Substitute this into the line equation [tex]\( y = 2x + k \)[/tex] with [tex]\( x = 2 \)[/tex] and [tex]\( y = 24 \)[/tex]:
[tex]\[ 24 = 2(2) + k \][/tex]
[tex]\[ 24 = 4 + k \][/tex]
[tex]\[ k = 20 \][/tex]
### Answer:
[tex]\[ k = 20 \][/tex]
These are the detailed solutions to each part of the question based on mathematical analysis.
### 2.1.1: Write down the equation of the asymptote of [tex]\( g \)[/tex].
The equation of the asymptote is a horizontal line that the graph approaches but never reaches. For the function [tex]\( g(x) = 2\left(\frac{1}{2}\right)^{x+1} - 4 \)[/tex], the horizontal asymptote is [tex]\( y = -4 \)[/tex].
### Answer:
[tex]\[ y = -4 \][/tex]
### 2.1.2: Write down the coordinates of D.
The turning point (vertex) of the quadratic function [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex] can be found by observing the standard form [tex]\( f(x) = a(x-h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex. Here, the vertex is at [tex]\((3, 25)\)[/tex].
### Answer:
[tex]\[ D = (3, 25) \][/tex]
### 2.1.3: Write down the range of [tex]\( f \)[/tex].
The function [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex] is a downward-opening parabola with a maximum value at the vertex [tex]\((3, 25)\)[/tex]. Therefore, the range of [tex]\( f \)[/tex] is [tex]\((-\infty, 25]\)[/tex].
### Answer:
[tex]\[ \text{Range of } f: (-\infty, 25] \][/tex]
### 2.1.4: Calculate the length of [tex]\( EB \)[/tex].
[tex]\( E \)[/tex] is the point where [tex]\( f \)[/tex] cuts the y-axis. Evaluating [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = -(0-3)^2 + 25 = -9 + 25 = 16 \][/tex]
So, [tex]\( E = (0, 16) \)[/tex].
[tex]\( B \)[/tex] is the point where [tex]\( g \)[/tex] cuts the y-axis. Evaluating [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = 2\left(\frac{1}{2}\right)^{0+1} - 4 = 2\left(\frac{1}{2}\right) - 4 = 1 - 4 = -3 \][/tex]
So, [tex]\( B = (0, -3) \)[/tex].
The length of [tex]\( EB \)[/tex] is the vertical distance between the y-coordinates of [tex]\( E \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ \text{Length of } EB = |16 - (-3)| = |16 + 3| = 19 \][/tex]
### Answer:
[tex]\[ \text{Length of } EB = 19 \][/tex]
### 2.1.5: Determine the values of [tex]\( x \)[/tex] for which [tex]\( f \)[/tex] is decreasing.
For the quadratic function [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex], it decreases when [tex]\( x \)[/tex] is to the right of the vertex. Since the vertex is at [tex]\( x = 3 \)[/tex], the function is decreasing for [tex]\( x > 3 \)[/tex].
### Answer:
[tex]\[ x \in (3, \infty) \][/tex]
### 2.1.6: Calculate the average gradient between points A and B.
Point [tex]\( A \)[/tex] is where [tex]\( f \)[/tex] cuts the x-axis. Solving [tex]\( f(x) = 0 \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ 0 = -(x-3)^2 + 25 \][/tex]
[tex]\[ (x-3)^2 = 25 \][/tex]
[tex]\[ x - 3= \pm 5 \][/tex]
Thus, the roots are [tex]\( x = 3 + 5 = 8 \)[/tex] and [tex]\( x = 3 - 5 = -2 \)[/tex]. If considering [tex]\( A \)[/tex] to be [tex]\( (8, 0) \)[/tex].
From part 2.1.4, [tex]\( B = (0, -3) \)[/tex].
The average gradient between [tex]\( A = (8, 0) \)[/tex] and [tex]\( B = (0, -3) \)[/tex]:
[tex]\[ \text{Average Gradient} = \frac{B_y - A_y}{B_x - A_x} = \frac{-3 - 0}{0 - 8} = \frac{-3}{-8} = 0.375 \][/tex]
### Answer:
[tex]\[ \text{Average Gradient} = 0.375 \][/tex]
### 2.1.7: If [tex]\( p(x) = f(x) + 2 \)[/tex], write down the coordinates of the turning point [tex]\( p \)[/tex].
The function [tex]\( p(x) = f(x) + 2 \)[/tex] is obtained by shifting [tex]\( f(x) \)[/tex] upward by 2 units. Therefore, the turning point [tex]\( D \)[/tex] of [tex]\( f \)[/tex] shifts to [tex]\( (3, 25 + 2) = (3, 27) \)[/tex].
### Answer:
[tex]\[ D_p = (3, 27) \][/tex]
### 2.1.8: Determine the value of [tex]\( k \)[/tex] for which the straight line [tex]\( y = 2x + k \)[/tex] will be a tangent to [tex]\( f \)[/tex].
To find [tex]\( k \)[/tex], we need the line [tex]\( y = 2x + k \)[/tex] to be tangent to [tex]\( f \)[/tex] at a single point. This happens where the slope of [tex]\( f \)[/tex] matches the slope of the line.
Given [tex]\( f(x) = -(x-3)^2 + 25 \)[/tex], the derivative [tex]\( f'(x) = -2(x-3) \)[/tex]. Equating this to the slope of the line (2):
[tex]\[ -2(x-3) = 2 \][/tex]
[tex]\[ -2x + 6 = 2 \][/tex]
[tex]\[ -2x = -4 \][/tex]
[tex]\[ x = 2 \][/tex]
Now, find [tex]\( f(2) \)[/tex]:
[tex]\[ f(2) = -(2-3)^2 + 25 = -1 + 25 = 24 \][/tex]
Substitute this into the line equation [tex]\( y = 2x + k \)[/tex] with [tex]\( x = 2 \)[/tex] and [tex]\( y = 24 \)[/tex]:
[tex]\[ 24 = 2(2) + k \][/tex]
[tex]\[ 24 = 4 + k \][/tex]
[tex]\[ k = 20 \][/tex]
### Answer:
[tex]\[ k = 20 \][/tex]
These are the detailed solutions to each part of the question based on mathematical analysis.
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