Let's solve each question step-by-step.
QUESTION 4
Given the quadratic pattern: 244, 193, 148, 109, ...
4.1 Write down the next term of the pattern. (2)
To find the next term, we first calculate the first differences:
- First difference between 244 and 193: \( 244 - 193 = 51 \)
- First difference between 193 and 148: \( 193 - 148 = 45 \)
- First difference between 148 and 109: \( 148 - 109 = 39 \)
Now let's calculate the second differences:
- Second difference between 51 and 45: \( 51 - 45 = 6 \)
- Second difference between 45 and 39: \( 45 - 39 = 6 \)
Since the second differences are constant (6), this confirms the pattern is quadratic.
To find the next first difference, we subtract 6 from the previous first difference:
- Next first difference: \( 39 - 6 = 33 \)
Now, subtract this from the last known term (109) to find the next term:
- Next term: \( 109 - 33 = 76 \)
Next term: 76
4.2 Determine a formula for the \( n \)th term of the pattern. (4)
The general form of a quadratic sequence is \[ a_n = An^2 + Bn + C \]
To find coefficients A, B, and C, we can form a system of equations using known terms:
-
\( a_1 = 244 \) implies: \[ A(1^2) + B(1) + C = 244 \] \[ A + B + C = 244 , (1) \]
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\( a_2 = 193 \) implies: \[ A(2^2) + B(2) + C = 193 \] \[ 4A + 2B + C = 193 , (2) \]
-
\( a_3 = 148 \) implies: \[ A(3^2) + B(3) + C = 148 \] \[ 9A + 3B + C = 148 , (3) \]
Now we can solve the system:
Subtract (1) from (2) and (2) from (3):
From (2) - (1): \[ 4A + 2B + C - (A + B + C) = 193 - 244 \] \[ 3A + B = -51 , (4) \]
From (3) - (2): \[ 9A + 3B + C - (4A + 2B + C) = 148 - 193 \] \[ 5A + B = -45 , (5) \]
Now subtract (4) from (5): \[ 5A + B - (3A + B) = -45 + 51 \] \[ 2A = 6 \] \[ A = 3 \]
Substituting \( A = 3 \) back into (4): \[ 3(3) + B = -51 \] \[ 9 + B = -51 \] \[ B = -60 \]
Substituting \( A \) and \( B \) back into (1): \[ 3 - 60 + C = 244 \] \[ C = 244 + 60 - 3 \] \[ C = 301 \]
Thus, the formula for the \( n \)th term is: \[ a_n = 3n^2 - 60n + 301 \]
Formula: \( a_n = 3n^2 - 60n + 301 \)
4.3 Which term of the pattern will have a value of 508? (4)
Set the \( n \)th term equal to 508: \[ 3n^2 - 60n + 301 = 508 \] \[ 3n^2 - 60n + 301 - 508 = 0 \] \[ 3n^2 - 60n - 207 = 0 \]
Now we can simplify by dividing through by 3: \[ n^2 - 20n - 69 = 0 \]
Using the quadratic formula: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{20 \pm \sqrt{(-20)^2 - 4(1)(-69)}}{2(1)}\] \[ n = \frac{20 \pm \sqrt{400 + 276}}{2} \] \[ n = \frac{20 \pm \sqrt{676}}{2} \] \[ n = \frac{20 \pm 26}{2} \]
Calculating the roots: \[ n_1 = \frac{46}{2} = 23 \] \[ n_2 = \frac{-6}{2} = -3 \] (not valid since n must be positive)
Thus, the term where \( a_n = 508 \) is \( n = 23 \).
Term: 23
4.4 Between which TWO consecutive terms of the quadratic pattern will the first difference be 453? (3)
Since the first difference is represented by the sequence: 51, 45, 39, 33, ...
The first difference decreases by 6 each time. The form of the first difference can be expressed as: \[ d_n = 51 - 6(n - 1) \]
To find when \( d_n = 453 \): \[ 51 - 6(n - 1) = 453 \] \[ -6(n - 1) = 453 - 51 \] \[ -6(n - 1) = 402 \] \[ n - 1 = -67 \] \[ n = -66 \] (not valid since n must be positive)
Thus, there will be no two consecutive terms where the first difference equals 453 as it's outside the range observed so far. The first difference is decreasing, and will never reach that value.
Consecutive Terms: None
4.5 Show that all the terms of the quadratic pattern are positive. (4)
To find when \( a_n \) is positive: \[ 3n^2 - 60n + 301 > 0 \]
Using the quadratic formula to find the roots: \[ n = \frac{-(-60) \pm \sqrt{(-60)^2 - 4(3)(301)}}{2(3)} \] \[ n = \frac{60 \pm \sqrt{3600 - 3604}}{6} = \frac{60 \pm \sqrt{-4}}{6} \] \[ n = \frac{60 \pm 2i}{6} \]
The discriminant is negative, indicating there are no real roots, meaning the quadratic never touches the x-axis. Since \( a_n \) opens upwards (A > 0), the function is always greater than zero.
Thus, all terms of the pattern are positive.
Show that: All terms are positive.
QUESTION 5
Given the hyperbola: \[ f(x) = \frac{3}{x - 2} + 1 \]
5.1 Write down the equations of the asymptotes of \( f \). (2)
The vertical asymptote occurs where the denominator is zero: \[ x - 2 = 0 \rightarrow x = 2 \]
The horizontal asymptote occurs as \( x \) approaches infinity or negative infinity. As \( x \rightarrow \infty, f(x) \rightarrow 1 \): \[ y = 1 \]
Asymptotes: \( x = 2 \) and \( y = 1 \)
5.2 Write down the \( y \)-intercept of \( f \). (2)
To find the \( y \)-intercept, substitute \( x = 0 \) into the function: \[ f(0) = \frac{3}{0 - 2} + 1 = \frac{3}{-2} + 1 = -\frac{3}{2} + 1 = -\frac{1}{2} \]
\( y \)-intercept: \(-\frac{1}{2}\)
5.3 Determine the \( x \)-intercept of \( f \). (2)
To find the \( x \)-intercept, set \( f(x) = 0 \): \[ 0 = \frac{3}{x - 2} + 1 \] \[ \frac{3}{x - 2} = -1 \] \[ 3 = -1(x - 2) \] \[ 3 = -x + 2 \] \[ x = -1 \]
\( x \)-intercept: \(-1\)
5.4 Sketch the graph of \( f \), showing ALL intercepts with the axes and any asymptotes. (5)
To sketch the graph of \( f \):
- Draw the vertical asymptote \( x = 2 \) and horizontal asymptote \( y = 1 \).
- Plot the \( y \)-intercept (0, -0.5) and \( x \)-intercept (-1, 0).
- The curve approaches the asymptotes but never touches them, showing the typical shape of a hyperbola.
5.5 Write down the domain of \( h \), if \( h(x) = f(x - 2) \). (2)
To find the domain of \( h(x) = f(x - 2) \), we note that the asymptote of \( f(x) \) at \( x = 2 \) will shift to \( x - 2 = 2 \): \[ x = 4 \]
Thus, the function is undefined at \( x = 4 \).
So, the domain is: \[ x \in \mathbb{R}, x \neq 4 \]
Domain: \( x \in \mathbb{R}, , x \neq 4 \)
This concludes the answers and explanations for Questions 4 and 5. If you have further questions or require clarification on any part, please let me know!
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