10. The graph of a linear polynomial passes through the points and . (i) Find the polynomial . (ii) Find the coordinates where the graph of cuts the axes. (iii) Draw the graph of and verify your answers.

Understand the Question A general linear polynomial is represented as , where . Since the graph passes through and , these coordinates satisfy and , giving a system of two linear equations in and . The graph of (or ): Cuts the -axis where . Cuts the -axis where . The graph of a linear polynomial is always a straight line. Part (i) Find the polynomial . Step 1 · Set up equations using the given points Let the linear polynomial be: Since the graph passes through : Since the graph passes through : Step 2 · Solve for and Subtract equation (1) from equation (2): Substitute into equation (1): Therefore, the polynomial is: Answer > (i) Part (ii) Find the coordinates where the graph of cuts the axes. Step 1 · Find the coordinates on the axes To find where the graph cuts the -axis, set : Thus, the graph cuts the -axis at . To find where the graph cuts the -axis, set : Thus, the graph cuts the -axis at . Answer > (ii) The graph cuts the -axis at and the -axis at . Part (iii) Draw the graph of and verify your answers. Step 1 · Plot the points and draw the straight line Table of points on the line : Plot the points and on the graph and connect them with a straight line. Extending the line shows that it intercepts: The -axis at The -axis at This graphically verifies the solutions obtained. Answer > (iii) The graph is a straight line passing through and , verifying the axis intercepts at and . Common Mistakes Swapping Axis Conditions: Finding the -intercept requires setting , while finding the -intercept requires setting . Confusing the two leads to inverted coordinates. Sign Errors in Linear Polynomial Form: When solving , forgetting to change the sign to give .

Introduction to Linear Polynomials | EOT

Question 10

*10. 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.

Solution

Understand the Question

A general linear polynomial is represented as $p(x) = ax + b$ , where $a \ne 0$ .
Since the graph passes through $(1, 5)$ and $(3, 11)$ , these coordinates satisfy $p(1) = 5$ and $p(3) = 11$ , giving a system of two linear equations in $a$ and $b$ .
The graph of $p(x)$ $p (x)$ (or $y = p(x)$ $y = p (x)$ ):
- Cuts the $x$ -axis where $y = p(x) = 0$ .
- Cuts the $y$ -axis where $x = 0$ .
The graph of a linear polynomial is always a straight line.

(i) Find the polynomial $p(x)$ .

Step 1 · Set up equations using the given points

Let the linear polynomial be: $p(x) = ax + b$

Since the graph passes through $(1, 5)$ :

\begin{aligned} p(1) &= a(1) + b = 5 \\ a + b &= 5 \quad \dots (1) \end{aligned}

Since the graph passes through $(3, 11)$ :

\begin{aligned} p(3) &= a(3) + b = 11 \\ 3a + b &= 11 \quad \dots (2) \end{aligned}

Step 2 · Solve for $a$ and $b$

Subtract equation (1) from equation (2):

\begin{aligned} (3a + b) - (a + b) &= 11 - 5 \\ 2a &= 6 \\ a &= 3 \end{aligned}

Substitute $a = 3$ into equation (1):

\begin{aligned} 3 + b &= 5 \\ b &= 5 - 3 \\ b &= 2 \end{aligned}

Therefore, the polynomial is: $p(x) = 3x + 2$

Answer

(i) $p(x) = 3x + 2$

(ii) Find the coordinates where the graph of $p(x)$ cuts the axes.

Step 1 · Find the coordinates on the axes

To find where the graph cuts the $x$ -axis, set $p(x) = 0$ :

\begin{aligned} 3x + 2 &= 0 \\ 3x &= -2 \\ x &= -\dfrac{2}{3} \end{aligned}

Thus, the graph cuts the $x$ -axis at $\left(-\dfrac{2}{3}, 0\right)$ .

To find where the graph cuts the $y$ -axis, set $x = 0$ :

\begin{aligned} y = p(0) &= 3(0) + 2 \\ &= 2 \end{aligned}

Thus, the graph cuts the $y$ -axis at $(0, 2)$ .

Answer

(ii) The graph cuts the $x$ -axis at $\left(-\dfrac{2}{3}, 0\right)$ and the $y$ -axis at $(0, 2)$ .

(iii) Draw the graph of $p(x)$ and verify your answers.

Step 1 · Plot the points and draw the straight line

Table of points on the line $y = 3x + 2$ :

\begin{array}{|c|c|c|c|c|} \hline x & -\dfrac{2}{3} & 0 & 1 & 3 \\ \hline y = p(x) & 0 & 2 & 5 & 11 \\ \hline \end{array}

Plot the points $(1, 5)$ and $(3, 11)$ on the graph and connect them with a straight line. Extending the line shows that it intercepts:

The $x$ -axis at $\left(-\dfrac{2}{3}, 0\right)$
The $y$ -axis at $(0, 2)$

This graphically verifies the solutions obtained.

Answer

(iii) The graph is a straight line passing through $(1, 5)$ and $(3, 11)$ , verifying the axis intercepts at $\left(-\dfrac{2}{3}, 0\right)$ and $(0, 2)$ .

Common Mistakes

Swapping Axis Conditions: Finding the $x$ -intercept requires setting $y = 0$ , while finding the $y$ -intercept requires setting $x = 0$ . Confusing the two leads to inverted coordinates.
Sign Errors in Linear Polynomial Form: When solving $3x + 2 = 0$ , forgetting to change the sign to give $x = -\dfrac{2}{3}$ .

Question 10

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