Detailed Answer

a)

\[ f(2) = 0\]

b) To find it, we need to use the property that the domain of the initial function is the range of the inverse, and vice versa. So:

\[ f(0) = 8\]

\[ f^{-1}(8) = 0\]

c) This will be a reflection in the y-axis.

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Detailed Answer

a) To obtain \( h(x) = 3(x+2)^2 \) from \( g(x) = x^2 \), the following geometric transformations are applied:

• Horizontal translation: Shift the graph 2 units in the negatvie x-direction.
• Vertical stretch: Stretch the graph by a factor of 3.

b) We have to follow the transformations outlined in part (a):

• First, shift the x-coordinate of \( P \) 2 units to the left: \( 2 - 2 = 0 \).
• Now apply the vertical stretch by multiplying the y-coordinate \( 3 \times 4 = 12 \).

Therefore, the coordinates of \( Q \) are \( (0, 12) \).

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Detailed Answer

a) (i), (ii), (iii) The transformations given for \( g(x) = \frac{1}{2}(f(x-2)) + 3 \) allow us to identify the coefficients:

• horizontal translation: From \( f(x-2) \), we see a horizontal translation of 2 units to the right. Therefore, \( a = 2 \).
• vertical translation: The constant term \( +3 \) indicates a vertical translation of 3 units upwards. Therefore, \( b = 3 \).
• vertical stretch: The coefficient \( \frac{1}{2} \) indicates a vertical stretch by a factor of \( \frac{1}{2} \). Therefore, \( c = \frac{1}{2} \).

So, \( a = 2 \), \( b = 3 \), and \( c = \frac{1}{2} \).

b) To find the coordinates of point \( B \) on \( h(x) = 4g(x) - 2 \), we apply the following steps:

• There are no transformation for the x-values.
• Let's apply transformation for the y-value: \( h(1) = 4 \times \frac{7}{2} - 2 = 14 - 2 = 12 \).

Therefore, the coordinates of point \( B \) are \( (1, 12) \).

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Detailed Answer

a) (i), (ii), (iii) To find the values of \( p \), \( q \), and \( r \) for \( g(x) = p(x+q)^3 + r \), we apply the given transformations to \( f(x) = 5x^3 + 25 \):

• i. (vertical stretch): The vertical stretch by \( \frac{1}{2} \) means the coefficient of \( x^3 \) is halved. Since \( f(x) \) has the term \( 5x^3 \), we have \( p = \frac{5}{2} \).
• ii. (horizontal translation): The horizontal translation is given by the vector \( \begin{pmatrix} -2 \\ 3 \end{pmatrix} \). The horizontal component \( -2 \) means the function is shifted 2 units to the right. Therefore, \( q = -2 \).
• iii. (vertical translation): The vertical component of the vector \( \begin{pmatrix} -2 \\ 3 \end{pmatrix} \) is \( 3 \), meaning the function is shifted 3 units up. Therefore, \( r = 3 \).

b) To find \( g'(x) \), we differentiate \( g(x) = \frac{5}{2}(x-2)^3 + 3 \):

\[ g'(x) = \frac{5}{2} \times 3(x-2)^2 = \frac{15}{2}(x-2)^2 \]

c) The function \( g(x) \) is increasing where \( g'(x) > 0 \). Since \( g'(x) = \frac{15}{2}(x-2)^2 \), which is always non-negative, \( g(x) \) is increasing for all \( x \in \mathbb{R} \).

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Detailed Answer

a)

\[ f(1) = (1)^2 + 2(1) - 8 = 1 + 2 - 8 = -5 \]

b) Finding the factors of \(-8\) that add up to \(2\) yields \(4\) and \(-2\).

\[ f(x) = (x + 4)(x - 2) \]

Thus, \( a = 4 \) and \( b = 2 \).

c) To find the distance traveled by the particle from \( x = 3 \) to \( x = 6 \), we calculate the integral:

\[ \text{Distance} = \int_{3}^{6} |x^2 + 2x - 8| \, dx = 66 \]

This can be easily found in the GDC:

d) Reflecting in the y-axis:

\[ f(-x) = (-x)^2 + 2(-x) - 8 = x^2 - 2x - 8 \]

Translate 2 units to the left: \[ g(x) = (x + 2)^2 - 2(x + 2) - 8 \]

\[ = x^2 + 4x + 4 - 2x - 4 - 8 \]

\[ = x^2 + 2x - 8 \]

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Detailed Answer

a)

• Horizontal Translation: The term \( x + \pi \) indicates a horizontal shift by \(-\pi\) units. This translates the function to the left by \(\pi\) units.

• Vertical Translation: The change from \( a \) to \( b \) shows a vertical shift. Since \( b \) is less than \( a \), the function is shifted downwards by \( a - b \) units.

b) (i) & (ii) Point \( A(0, 4) \) is on \( f(x) \), so:

\[ f(0) = 2\sin(0) + a = a \] We need to find \( a \) such that when \( x = 0 \), \( f(x) = 4 \). Hence, \( a = 4 \).

Point \( B(\pi, 2) \) is on \( g(x) \), so: \[ g(\pi) = 2\sin(\pi + \pi) + b = 2\sin(2\pi) + b = b \] Since \( g(\pi) = 2 \), we have \( b = 2 \).

c) The function \( g(x) = 2\sin(x + \pi) + b \) has the same amplitude and period as \( f(x) \), but with a vertical shift. The range of \( \sin(x) \) is \([-1, 1]\), so the range of \( 2\sin(x) \) is \([-2, 2]\). Adding \( b = 2 \), the range of \( g(x) \) is:

\[ [-2 + 2, 2 + 2] = [0, 4] \]

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Detailed Answer

a) To find \( a \), we use the axis of symmetry formula for a quadratic function \( f(x) = 3x^2 + ax - 18 \). The axis of symmetry is given by:

\[ x = -\frac{b}{2a} \]

For the function \( f(x) = 3x^2 + ax - 18 \), the axis of symmetry is \( \frac{1}{2} \). Thus:

\[ \frac{1}{2} = -\frac{a}{2 \times 3} \]

Solving for \( a \):

\[ \frac{1}{2} = -\frac{a}{6} \]

\[ a = -3 \]

b) Substitute \( x = \frac{1}{2} \) into \( f(x) \):

\[ f\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) - 18 \]

\[ = 3 \cdot \frac{1}{4} - \frac{3}{2} - 18 = \frac{3}{4} - \frac{3}{2} - 18 = -\frac{63}{4} \]

Thus, the vertex is: \(\left(\frac{1}{2}, -\frac{63}{4}\right)\)

c) To find the exact roots of the function, we solve:

\[ 3x^2 - 3x - 18 = 0 \]

\[ x^2 - x - 6 = 0 \]

\[ (x-3)(x+2) = 0 \]

\[ x = 3, \ x = -2 \]

d) Point \( M(1, -18) \) on \( f(x) \) is reflected across the x-axis to get \( N \) on \( g(x) \). The reflection changes the y-coordinate sign and doesn't affect the x-coordinate:

\[ N(1, 18) \]

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Detailed Answer

a) (i) & (ii) Vertical asymptote occurs where the denominator of the fraction equals zero, which makes the function undefined:

\[ x + 1 = 0 \]

\[ x = -1 \]

As \( x \) approaches infinity, the term \(\frac{2}{x+1}\) approaches zero. Thus, the function approaches:

\[ f(x) \to 2 - 0 = 2 \]

The horizontal asymptote is: \( y = 2 \)

b) (i) Set \( f(x) = 0 \):

\[ 2 - \frac{2}{x+1} = 0 \]

\[ \frac{2}{x+1} = 2 \]

\[ 2 = 2(x + 1) \]

\[ 2 = 2x + 2 \]

\[ 0 = 2x \]

\[ x = 0 \]

The x-intercept is: \( (0, 0) \)

b) (ii) Set \( x=0 \):

\[ f(0) = 2 - \frac{2}{0+1} \]

\[ f(0) = 2 - 2 = 0 \]

The y-intercept is: \( (0, 0) \)

c) Let's start with the reflection in the y-axis. Replace \( x \) with \( -x \) in \( f(x) \):

\[ f(-x) = 2 - \frac{2}{-x + 1} \]

\[ = 2 + \frac{2}{x - 1} \]

For the horizontal stretch of 3 we replace \( x \) with \( \frac{x}{3} \):

\[ g(x) = 2 + \frac{2}{\frac{x}{3} - 1} \]

\[ g(x) = 2 + \frac{2}{\frac{x - 3}{3}} \]

\[ g(x) = 2 + \frac{6}{x - 3} \]

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