QUESTION 5 Given: $$ f(x)=\frac{1}{x-3}-\frac{2 x+6}{x+3} $$ 5.1 Show that $$ f(x) $$ can be written as $$ f(x)=\frac{1}{x-3}-2 $$ 5.2 Write down the equations of the asymptotes of $$ f $$. (2) 5.3 Determine the $$ x $$-intercept of $$ f $$. (3) 5.4 Determine the $$ y $$-intercept of $$ f $$. (2) 5.5 Sketch the graph of $$ f $$. Show clearly ALL the intercepts with the axes and the asymptotes. 5.6 Determine the equation of the axis of symmetry of $$ f $$ having positive gradient. (3) (3) 5.7 The graph of $$ f $$ is transformed to obtain the graph of $$ h(x)=\frac{1}{x} $$. Describe the transformation from $$ f $$ to $$ h $$. (2) 5.8 Write down the domain of $$ h $$. (2) [19]

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We are given 
  f(x) = 1⁄(x – 3) – (2x + 6)⁄(x + 3).

In answering the sub‐questions note that when we “simplify” f(x), we must remember that the original function is not defined when a denominator is 0; that is, x ≠ 3 and x ≠ –3. (Later you’ll see that although the simplified form is 1⁄(x–3) – 2, the hole at x = –3 remains.)

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5.1 Show that f(x) can be written as f(x) = 1⁄(x – 3) – 2

Notice that in the second term the numerator factors neatly:
  2x + 6 = 2(x + 3).

Thus

f(x) = 1⁄(x – 3) – [2(x + 3)]⁄(x + 3).

Provided x ≠ –3 we can cancel (x + 3) in the second term so that

f(x) = 1⁄(x – 3) – 2.

So, f(x) = 1⁄(x – 3) – 2 for all x with x ≠ 3 and x ≠ –3.

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5.2 Write down the equations of the asymptotes of f.

There are two types of asymptotes to consider: vertical and horizontal.

• Vertical asymptote comes from where a denominator is 0 in the simplified form. In f(x) = 1⁄(x – 3) – 2 the only problematic denominator is x – 3, so we have
  x – 3 = 0 ⟹ x = 3.

• Horizontal asymptote: As x → ±∞, 1⁄(x – 3) → 0 so f(x) → –2.
  Thus the horizontal asymptote is y = –2.

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5.3 Determine the x-intercept of f.

The x-intercept(s) are found by setting f(x) = 0.

Starting with the simplified form:
  1⁄(x – 3) – 2 = 0  ⟹  1⁄(x – 3) = 2.

Multiply both sides by (x – 3):
  1 = 2(x – 3)
  ⟹ 2(x – 3) = 1  ⟹  x – 3 = 1⁄2  ⟹  x = 3 + 1⁄2 = 7⁄2.

Thus the x-intercept is the point (7⁄2, 0).

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5.4 Determine the y-intercept of f.

The y-intercept is found by letting x = 0 (provided 0 is in the domain; note 0 ≠ 3 and 0 ≠ –3 so it is valid).

Using the simplified form:
  f(0) = 1⁄(0 – 3) – 2 = 1⁄(–3) – 2 = –1⁄3 – 2 = –1⁄3 – 6⁄3 = –7⁄3.

Thus the y-intercept is (0, –7⁄3).

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5.5 Sketch the graph of f. Show clearly ALL the intercepts with the axes and the asymptotes.

Key features of the graph of f(x) = 1⁄(x – 3) – 2 are as follows:

1. Vertical asymptote: x = 3.
2. Horizontal asymptote: y = –2.
3. x-intercept: (7⁄2, 0).
4. y-intercept: (0, –7⁄3).
5. Domain: Although the simplified form is valid when x ≠ 3, recall from the original form that x = –3 must also be excluded. At x = –3 the “simplification” has removed a factor; hence, there is a removable discontinuity (a hole). To find its location note that the simplified expression gives
  f(–3) = 1⁄(–3 – 3) – 2 = 1⁄(–6) – 2 = –1⁄6 – 2 = –13⁄6.
So, there is a hole at (–3, –13⁄6).

When sketching the graph, be sure to:
 • Draw the dashed vertical asymptote x = 3 and the dashed horizontal asymptote y = –2.
 • Plot the intercepts: (7⁄2, 0) and (0, –7⁄3).
 • Mark the hole at (–3, –13⁄6) with an open circle.
 • Draw the curve approaching the asymptotes as x → 3± and as x → ±∞.

[An accurate sketch would show the branch for x < 3 (which also includes the hole at x = –3) and the branch for x > 3.]

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5.6 Determine the equation of the axis of symmetry of f having positive gradient.

Although the function f(x) = 1⁄(x – 3) – 2 looks like a “shifted” reciprocal function, it is equivalent to writing
  y + 2 = 1⁄(x – 3)
or, equivalently,
  (x – 3)(y + 2) = 1.

Introducing new variables
  u = x – 3  and  v = y + 2,
this equation becomes
  u v = 1,
which is the equation of a rectangular hyperbola. A rectangular hyperbola has two axes of symmetry in the u–v plane – these are the lines
  v = u  and  v = –u.
Now, transforming these back into x and y:

1. For v = u:
  y + 2 = x – 3  ⟹  y = x – 5.

2. For v = –u:
  y + 2 = –(x – 3)  ⟹  y + 2 = –x + 3  ⟹  y = –x + 1.

Since we are asked for the axis of symmetry having positive gradient, we choose the line with gradient 1:
  y = x – 5.

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5.7 The graph of f is transformed to obtain the graph of h(x) = 1⁄x. Describe the transformation from f to h.

We have
  f(x) = 1⁄(x – 3) – 2  and  h(x) = 1⁄x.
To go from f(x) to h(x), notice that if we “reverse” the translation that appears in f it will yield h:
 • Start with f(x) = 1⁄(x – 3) – 2.
 • Shift the graph 3 units to the left. Let u = x – 3 so that
  g(u) = 1⁄u – 2.
 • Now shift the graph 2 units up:
  h(u) = [1⁄u – 2] + 2 = 1⁄u.
Finally, renaming u as x gives h(x) = 1⁄x.

Thus the transformation is: shift the graph of f 3 units to the left and then 2 units up.

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5.8 Write down the domain of h.

The function h(x) = 1⁄x is defined for all real numbers except where the denominator is 0. Therefore, the domain of
The domain of $$ h(x) $$ is all real numbers except $$ x = 0 $$.

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