18N.2.SL.TZ0.S_3

pestleMathematicsAASLPaper 218N· sl-2-8-reciprocal-and-simple-rational-functions-equations-of-asymptotessource ↗

Let $f\left( x \right) = \frac{{6x - 1}}{{2x + 3}}$ , for $x \ne - \frac{3}{2}$ .

For the graph of $f$ , find the $y$ -intercept.

[2]

a.i.

Hence or otherwise, write down $\mathop {{\text{lim}}}\limits_{x \to \infty } \left( {\frac{{6x - 1}}{{2x + 3}}} \right)$ .

[2]

Markscheme / solution

valid method (M1)

eg $f$ (0), sketch of graph

$y$ -intercept is $- \frac{1}{3}$ (exact), −0.333, $\left( {0,\, - \frac{1}{3}} \right)$ A1 N2

[2 marks]

a.i.

valid approach (M1)

eg recognizing that $\mathop {{\text{lim}}}\limits_{x \to \infty } f\left( x \right)$ is related to the horizontal asymptote,

table with large values of $x$ , their $y$ value from (a)(iii), L’Hopital’s rule $\mathop {{\text{lim}}}\limits_{x \to \infty } f\left( x \right) = 3$ .

$\mathop {{\text{lim}}}\limits_{x \to \infty } \left( {\frac{{6x - 1}}{{2x + 3}}} \right) = 3$ A1 N2

[2 marks]

Examiners’ report

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a.i.

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