Detailed Answer

a) Knowing that the vertical asymptote is at \( x = 3 \), we have that:

\[ x + a = 0 \implies a = -3 \]

b) The formula for the horizontal asymptote of the function is:

\[ y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} = \frac{3}{1} = 3 \]

c) To find the x-intercept, set \( f(x) = 0 \):

\[ 3x + 2 = 0 \implies x = -\frac{2}{3} \]

d) To find the y-intercept, set \( x = 0 \):

\[ f(0) = \frac{3(0) + 2}{0 - 3} = \frac{2}{-3} = -\frac{2}{3} \]

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Detailed Answer

a) To find \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \):

\[ f(x) = \frac{2x}{x+1}, \quad g(x) = x + 3 \]

Thus, \( (f \circ g)(x) = f(g(x)) \):

\[ f(g(x)) = f(x + 3) = \frac{2(x + 3)}{(x + 3) + 1} = \frac{2(x + 3)}{x + 4} = \frac{2x + 6}{x + 4} \]

b) To find the vertical asymptote of \( (f \circ g)(x) \), we set the denominator to zero:

\[ x + 4 = 0 \implies x = -4 \]

c) The leading coefficient of the numerator (2x) is 2, and the leading coefficient of the denominator (x) is 1. Thus, the horizontal asymptote is:

\[ y = \frac{2}{1} = 2 \]

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Detailed Answer

a) We can see that the vertical asymptote is at \( x = 3 \), meaning that \( b = 3 \).

b) The horizontal asymptote is at \( y = 5 \), meaning that the ratio of the leading coefficients of the numerator and the denominator has to be equal to 5. The coefficient of the denominator is 1, meaning that \( a = 5 \).

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Detailed Answer

a)

\[ 3x - 1 = 0 \]

\[ 3x = 1 \]

\[ x = \frac{1}{3} \]

b)

\[ f(0) = \frac{3(0) - 1}{2(0) + 2} = -\frac{1}{2} \]

c) To find the equation of the vertical asymptote, set the denominator equal to zero:

\[ 2x + 2 = 0 \]

\[ 2x = -2 \]

\[ x = -1 \]

So, the vertical asymptote is \( x = -1 \).

d)

To find the equation of the horizontal asymptote, compare the degrees of numerator and denominator:

Since both are linear (degree 1), the horizontal asymptote is determined by the ratio of their leading coefficients:

\[ y = \frac{3}{2} \]

So, the horizontal asymptote is \( y = \frac{3}{2} \).

e)

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Detailed Answer

a) (i)

\[ 7x + 2 = 0 \implies 7x = -2 \implies x = -\frac{2}{7} \]

a) (ii)

\[ y = \frac{7(0) + 2}{6(0) - 8} = \frac{2}{-8} = -\frac{1}{4} \]

b) To find the inverse function, swap \( x \) and \( y \) and solve for \( y \):

\[ x = \frac{7y + 2}{6y - 8} \]

\[ x(6y - 8) = 7y + 2 \]

\[ 6xy - 8x = 7y + 2 \implies 6xy - 7y = 8x + 2 \implies y(6x - 7) = 8x + 2 \implies y = \frac{8x + 2}{6x - 7} \]

So, the inverse function is:

\[ f^{-1}(x) = \frac{8x + 2}{6x - 7} \]

c) By plotting the functions we can easily find the points of intersection:

d) First, we have to find the gradient of the line AB:

\[ m = \frac{2.45 - (-0.123)}{2.45 - (-0.123)} = 1 \]

So the gradient of the perpendicular line has to be equal to -1. Now, knowing that \( L_1 \) passess through \( P(4,4) \), we can use the formula for the equation of the line:

\[ y - 4 = -1(x - 4)\]

\[ y = -x + 8\]

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Detailed Answer

a)

Let's put both points into the function:

For point \( P(1, 7) \):

\[ 7 = a + \frac{b}{2(1) + 3} \]

\[ 7 = a + \frac{b}{5} \tag{1} \]

For point \( Q\left(\frac{7}{2}, 6\right) \):

\[ 6 = a + \frac{b}{2\left(\frac{7}{2}\right) + 3} \]

\[ 6 = a + \frac{b}{7 + 3} \]

\[ 6 = a + \frac{b}{10} \tag{2} \]

Subtract equation (2) from equation (1) to eliminate \( a \):

\[ 7 - 6 = \left(a + \frac{b}{5}\right) - \left(a + \frac{b}{10}\right) \]

\[ 1 = \frac{2b - b}{10} \]

\[ 1 = \frac{b}{10} \]

\[ b = 10 \]

Substitute \( b = 10 \) back into equation (2) to solve for \( a \):

\[ 6 = a + \frac{10}{10} \]

\[ 6 = a + 1 \]

\[ a = 5 \]

So, the values are:

\[ a = 5 \]

\[ b = 10 \]

b)

\[ \lim_{x \to \infty} \left(a + \frac{b}{2x + 3}\right) \]

As \( x \) approaches infinity, \( \frac{b}{2x + 3} \) approaches 0, so the limit is \( a \):

\[ \lim_{x \to \infty} f(x) = a \]

Given \( a = 5 \):

\[ \lim_{x \to \infty} f(x) = 5 \]

So, the value of the limit is 5.

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Detailed Answer

a)

\[ (g \circ f)(x) = g(f(x)) = g(x - 4) \]

\[ g(x - 4) = \frac{2 + (x - 4)}{3(x - 4) - 1} \]

\[ = \frac{x - 2}{3x - 13} \]

b)

To find the inverse \( (g \circ f)^{-1}(x) \), start with \( y = \frac{x - 2}{3x - 13} \) and solve for \( x \) in terms of \( y \):

\[ x(3y - 13) = y - 2 \]

\[ 3yx - 13x = y - 2 \]

\[ 2 - 13x = y - 3yx \]

\[ 2 - 13x = y(1-3x) \]

\[ y = \frac{2-13x}{1-3x} \]

Thus, we have shown that they are indeed equal.

c)

The leading coefficients are \(-13\) and \(-3\):

\[ y = \frac{-13}{-3} = \frac{13}{3} \]

So, the horizontal asymptote is \( y = \frac{13}{3} \).

d) By plotting two functions and finding the intersects, we find that:

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Detailed Answer

a)

\[ f(0) = \frac{3(0) - 2}{2(0) + 3} = \frac{-2}{3} \]

b)

For the horizontal asymptote of \( g(x) = \frac{4x + 2}{2x - 1} \), we look at the leading coefficients:

The leading coefficients of the numerator and the denominator are 4 and 2, respectively. Thus:

\[ y = \frac{4}{2} = 2 \]

So, the horizontal asymptote is \( y = 2 \).

c)

To find the inverse we first replace \( x \) with \( y \):

\[ x = \frac{3y + 2}{2y + 3} \]

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

\[ 2xy + 3x = 3y - 2 \]

\[ 2 + 3x = 3y - 2xy \]

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

\[ y = \frac{3x+2}{3-2x} \]

Thus, we have shown that the inverse is \( f^{-1}(x) = \frac{3x + 2}{3 - 2x} \).

d) To find \( (f \circ g)(x) \), we start with \( f(x) \) and \( g(x) \):

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

\[ g(x) = \frac{4x + 2}{2x - 1} \]

Then:

\[ (f \circ g)(x) = f(g(x)) \]

\[ = f\left( \frac{4x + 2}{2x - 1} \right) \]

Substitute \( g(x) \) into \( f(x) \):

\[ f\left( \frac{4x + 2}{2x - 1} \right) = \frac{3\left( \frac{4x + 2}{2x - 1} \right) - 2}{2\left( \frac{4x + 2}{2x - 1} \right) + 3} \]

Simplify the numerator:

\[ 3\left( \frac{4x + 2}{2x - 1} \right) - 2 = \frac{12x + 6}{2x - 1} - 2 = \frac{12x + 6 - 4x + 2}{2x - 1} = \frac{8x + 8}{2x - 1} \]

Simplify the denominator:

\[ 2\left( \frac{4x + 2}{2x - 1} \right) + 3 = \frac{8x + 4}{2x - 1} + 3 = \frac{8x + 4 + 6x - 3}{2x - 1} = \frac{14x + 1}{2x - 1} \]

Therefore:

\[ f(g(x)) = \frac{\frac{8x + 8}{2x - 1}}{\frac{14x + 1}{2x - 1}} = \frac{8x + 8}{14x + 1} = \frac{8(x + 1)}{14x + 1} \]

e)To solve it, we have to find the points of interesection by plotting both functions:

So, from the figure above we can see that point \( A \) has coordinates \( (-0.18, -0.96) \) and point \( B \) has coordinates \( (-11.32, 1.83) \).

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