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Let $f(x) = p + \\frac{3}{x-q}, x \\ne $ {q}.

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Parts (a) and (b) both relate to the vertical asymptote. This occurs when the denominator in the fraction $x - ${q} = 0

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Therefore q = {q} and the vertical asymptote is at $x = $ {q}

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Part (c) is about the y-intercept. This is when $x = 0$

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$f(0) = p + \\frac{3}{-3}$

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$f(0) = p - 1$

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$p - 1= $ {p-1}

\n

$p= $ {p}

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q in function, determines vertical asymptote.

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p in function, determines y-intercept

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Write down the value of $q$

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Write down the value of the vertical asymptote.

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The graph intercepts the y-axis at the point (0,{p-1})

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Write down the value of $p$

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