A sketch of the hyperbola $f(x) = \frac{d - x}{x - p}$, where $p$ and $d$ are constants, is given below. The dotted lines are the asymptotes. The point $A(5; 0)$ is given on the graph of $f$. 7.1 Determine the values of $d$ and $p$. 7.2 Show that the equation can be written as $y = \frac{3}{x - 2} - 1$. 7.3 Write down the image of $A$ if $A$ is reflected about the axis of symmetry $y = x - 3$.

NSC Mathematics Functions: A sketch of the hyperbola $f(x

A sketch of the hyperbola $f(x) = \frac{d - x}{x - p}$, where $p$ and $d$ are constants, is given below - NSC Mathematics - Question 7 - 2017 - Paper 1

Question 7

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A sketch of the hyperbola $f(x) = \frac{d - x}{x - p}$, where $p$ and $d$ are constants, is given below. The dotted lines are the asymptotes. The point $A(5; 0)$ ... show full transcript

Worked Solution & Example Answer:A sketch of the hyperbola $f(x) = \frac{d - x}{x - p}$, where $p$ and $d$ are constants, is given below - NSC Mathematics - Question 7 - 2017 - Paper 1

Step 1

7.1 Determine the values of d and p.

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Answer

To find the values of $d$ and $p$ , we start by observing the hyperbola's asymptotes. The given equation is:

$f(x) = \frac{d - x}{x - p}$

As $x \to \infty$ , the horizontal asymptote will be determined by the degrees of the polynomial in the numerator and the denominator. Hence, we have:

For vertical asymptote $x = p$ where the function is undefined, and
The horizontal asymptote will be given by the ratio of leading coefficients which leads to $d = 5$ and $p = 2$ .

Step 2

7.2 Show that the equation can be written as y = \frac{3}{x - 2} - 1.

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Answer

We can rewrite the hyperbola equation:

Starting from: $y = \frac{d - x}{x - p}$

Substituting the determined values: $y = \frac{5 - x}{x - 2}$

Rearranging this gives: $y = \frac{-(x - 5)}{x - 2} = -1 + \frac{3}{x - 2}$ Thus, we can express it as: $y = \frac{3}{x - 2} - 1$ .

Step 3

7.3 Write down the image of A if A is reflected about the axis of symmetry y = x - 3.

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Answer

To reflect point $A(5; 0)$ about the line $y = x - 3$ , we first identify the perpendicular line from $A$ to the axis of symmetry. The slope of the line $y = x - 3$ is 1, so the slope of the perpendicular line is -1.

The equation of the line that passes through $A(5, 0)$ can be expressed as: $y - 0 = -1(x - 5)$ which simplifies to: $y = -x + 5$

Setting this equal to the line of symmetry: $-x + 5 = x - 3$ Solving gives: $2x = 8 \Rightarrow x = 4$ Substituting back: $y = 4 - 3 = 1$

The midpoint between $A(5; 0)$ and its image $A'(x; y)$ must lie on the line of symmetry. Therefore, the reflected point is $A'(4 - 3; 1 + 3) = A(3; 2)$ .

A sketch of the hyperbola $f(x) = \frac{d - x}{x - p}$, where $p$ and $d$ are constants, is given below - NSC Mathematics - Question 7 - 2017 - Paper 1

Worked Solution & Example Answer:A sketch of the hyperbola $f(x) = \frac{d - x}{x - p}$, where $p$ and $d$ are constants, is given below - NSC Mathematics - Question 7 - 2017 - Paper 1

7.1 Determine the values of d and p.

7.2 Show that the equation can be written as y = \frac{3}{x - 2} - 1.

7.3 Write down the image of A if A is reflected about the axis of symmetry y = x - 3.

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