The graph of a linear polynomial \(p(x)\) passes through the points (1,5) and (3,11).
i) Find the polynomial p(x).
ii) Find the coordinates where the graph of p(x) cuts the axes.
iii) Draw the graph of p(x) and verify your answers.

Since \(p(x)\) is linear, we can write \(p(x) = ax + b\), with constants \(a\) and \(b\).
\(p(x)\) passes through the points \((1,5)\) and \((3,11)\).
So,
\(p(1) = 5\) ⇒ \(a(1) + b = 5\) ⇒ \(a + b = 5\) --- (I)
\(p(3) = 11\) ⇒ \(a(3) + b = 11\) ⇒ \(3a + b = 11\) ---(II)
Subtracting (I) from (II), we get
\((3a + b) − (a + b) = 11 − 5 \)
⇒ \(3a − a + b − b) = 6 \)
⇒ \(2a = 6 \)
⇒ \(a = 3 \)
Substituting this into equation (I), we get
\(3 + b = 5\) ⇒ \(b = 2\)

Thus \(p(x) = 3x + 2\)

For \(y-\) intercept, set \(x = 0\) ⇒ \(p(0) = 3(0) + 2 = 2\)
The line cuts the \(y-\) axis at \((0,2)\)
For \(x-\) intercept, set \(p(x) = 0\) ⇒ \( 3x + 2 = 0\) ⇒ \( x = \frac{−2}{3} \)
The line cuts the \(x-\) axis at \((\frac{−2}{3},0)\)