IB DP Maths Topic 2.5 Vertical and horizontal asymptotes SL Paper 2

 

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Question

Let \(f(x) = \frac{{3x}}{{x – q}}\), where \(x \ne q\).

Write down the equations of the vertical and horizontal asymptotes of the graph of \(f\).

[2]
a.

The vertical and horizontal asymptotes to the graph of \(f\) intersect at the point \({\text{Q}}(1,3)\).

Find the value of \(q\).

[2]
b.

The vertical and horizontal asymptotes to the graph of \(f\) intersect at the point \({\text{Q}}(1,3)\).

The point \({\text{P}}(x,{\text{ }}y)\) lies on the graph of \(f\). Show that \({\text{PQ}} = \sqrt {{{(x – 1)}^2} + {{\left( {\frac{3}{{x – 1}}} \right)}^2}} \).

[4]
c.

The vertical and horizontal asymptotes to the graph of \(f\) intersect at the point \({\text{Q}}(1,3)\).

Hence find the coordinates of the points on the graph of \(f\) that are closest to \((1,3)\).

[6]
d.
Answer/Explanation

Markscheme

\(x = q,{\text{ }}y = 3\)   (must be equations)     A1A1     N2

[2 marks]

a.

recognizing connection between point of intersection and asymptote     (R1)

eg     \(x = 1\)

\(q = 1\)     A1     N2

[2 marks]

b.

correct substitution into distance formula     A1

eg     \(\sqrt {{{(x – 1)}^2} + {{(y – 3)}^2}} \)

attempt to substitute \(y = \frac{{3x}}{{x – 1}}\)     (M1)

eg     \(\sqrt {{{(x – 1)}^2} + {{\left( {\frac{{3x}}{{x – 1}} – 3} \right)}^2}} \)

correct simplification of \(\left( {\frac{{3x}}{{x – 1}} – 3} \right)\)     (A1)

eg     \(\frac{{3x – 3x(x – 1)}}{{x – 1}}\)

correct expression clearly leading to the required answer     A1

eg     \(\frac{{3x – 3x + 3}}{{x – 1}},{\text{ }}\sqrt {{{(x – 1)}^2} + {{\left( {\frac{{3x – 3x + 3}}{{x – 1}}} \right)}^2}} \)

\({\text{PQ}} = \sqrt {{{(x – 1)}^2} + {{\left( {\frac{3}{{x – 1}}} \right)}^2}} \)     AG     N0

[4 marks]

c.

recognizing that closest is when \({\text{PQ}}\) is a minimum     (R1)

eg     sketch of \({\text{PQ}}\), \(({\text{PQ}})'(x) = 0\)

\(x =  – 0.73205{\text{ }}x = 2.73205\)   (seen anywhere)     A1A1

attempt to find y-coordinates     (M1)

eg     \(f( – 0.73205)\)

\((-0.73205, 1.267949) , (2.73205, 4.73205)\)

\((-0.732, 1.27) , (2.73, 4.73) \)    A1A1     N4

[6 marks]

d.

Question

Let \(f\left( x \right) = \frac{{8x – 5}}{{cx + 6}}\) for \(x \ne  – \frac{6}{c},\,\,c \ne 0\).

The line x = 3 is a vertical asymptote to the graph of f. Find the value of c.

[2]
a.

Write down the equation of the horizontal asymptote to the graph of f.

[2]
b.

The line y = k, where \(k \in \mathbb{R}\) intersects the graph of \(\left| {f\left( x \right)} \right|\) at exactly one point. Find the possible values of k.

[3]
c.
Answer/Explanation

Markscheme

valid approach       (M1)
eg  \(cx + 6 = 0,\,\, – \frac{6}{c} = 3\)

c = −2      A1 N2

[2 marks]

a.

valid approach (M1)
eg  \(\mathop {{\text{lim}}\,f}\limits_{x \to \infty } \left( x \right),\,\,y = \frac{8}{c}\)

y = −4 (must be an equation)      A1 N2

[2 marks]

b.

valid approach to analyze modulus function      (M1)
eg   sketch, horizontal asymptote at y = 4, y = 0

k = 4, k = 0     A2 N3

[3 marks]

c.
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