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Question 0113
Equations and Inequalities
2013 Paper 1 Question 2 Variant
Question
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It is given that
y
=
x
2
−
4
x
−
6
x
+
3
,
x
∈
R
,
x
≠
−
3.
y = \frac{ x^2 - 4 x - 6 }{ x + 3 }, \quad x \in \mathbb{R}, x \neq -3.
y
=
x
+
3
x
2
−
4
x
−
6
,
x
∈
R
,
x
=
−
3.
Without using a calculator, find the set of values that
y
{y}
y
can take.
[5]
Answer
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