Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
i) The graph of p(x) passes through the points (2,3) and (6,11).
ii) The graph of q(x) passes through the point (4,−1).
iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the points where these lines meet the x-axis.

(i) \(p(x)\) passes through \((2,3)\) and \((6,11)\):
\(3 = a(2) + b\)
⇒ \(2a + b = 3\) ---(I)
And
\(11 = a(6) + b\)
⇒ \(6a + b = 11\) ---(II)

Subtracting (I) from (II), we get
\( (6a + b) − (2a + b) = 11 − 3\)
⇒ \( (6 − 2)a + (b − b) = 8\)
⇒ \( 4a = 8\)
⇒ \( a = 2\)
Substituting \(a = 2\) in (I), we get
\(2(2) + b = 3\)
⇒ \( 4 + b = 3\)
⇒ \( b = −1\)

Thus \(p(x) = 2x − 1\)

Since \(q(x)\) is parallel to \(p(x)\), slope of \(q(x)\) is also \(2\), i.e., \(c = 2\). So,
\(q(x) = 2x + d\)
(ii) \(q(x)\) passes through \((4,−1\)):
\(−1 = 2(4) + d\)
⇒ \(−1 = 8 + d\)
⇒ \(d = −9 \)
Thus \(q(x) = 2x − 9\)

On the \(x-\)axis, \(y = 0\)
For \(p(x)\):
\(0 = 2x − 1\)
⇒ \(2x = 1 \)
⇒ \(x =\frac{1}{2} \)
\(p(x)\) meets the \(x-\)axis at (\(\frac{1}{2},0\))

For \(q(x)\):
\(0 = 2x − 9\)
⇒ \(2x = 9 \)
⇒ \(x =\frac{9}{2} \)
\(q(x)\) meets the \(x-\)axis at (\(\frac{9}{2},0\))