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 linear polynomial has the general form , where . The graph of a linear polynomial is a straight line representing . Since the line passes through and , substituting these coordinates gives a system of linear equations in and . The line intersects the -axis where , and the -axis where . Part (i) Find the polynomial . Step 1 · Set Up the Linear Equations Let the linear polynomial be: Since the graph passes through and : Step 2 · Solve for and Subtract equation from equation : Substitute into equation : Therefore, the polynomial is: Answer > (i) Part (ii) Find the coordinates where the graph of cuts the axes. Step 1 · Find the Intersection with the -axis On the -axis, : So, the graph cuts the -axis at . Step 2 · Find the Intersection with the -axis On the -axis, : So, 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 Line To graph , we list the key points: Plot the points , , , and on the coordinate plane and join them with a straight line. Step 2 · Verify the Solutions From the plotted graph: 1. The line passes through and , verifying the polynomial . 2. The line intersects the -axis at and the -axis at , verifying the intercept coordinates. Answer > (iii) The straight line graph of passes through and and intersects the axes at and , verifying all calculated values. Common Mistakes Swapping Axis Conditions: Confusing for the -intercept and for the -intercept. Always remember that the -axis has and the -axis has . Sign Error in Intercept: Solving as instead of .

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 linear polynomial has the general form $p(x) = ax + b$ , where $a \ne 0$ .
The graph of a linear polynomial is a straight line representing $y = p(x) = ax + b$ .
Since the line passes through $(1, 5)$ and $(3, 11)$ , substituting these coordinates gives a system of linear equations in $a$ and $b$ .
The line intersects the $x$ -axis where $y = 0$ , and the $y$ -axis where $x = 0$ .

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

Step 1 · Set Up the Linear Equations

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

Since the graph passes through $(1, 5)$ and $(3, 11)$ :

\begin{aligned} p(1) = a(1) + b = 5 &\implies a + b = 5 \quad \dots (1) \\[0.6em] p(3) = a(3) + b = 11 &\implies 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 &= 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 Intersection with the $x$ -axis

On the $x$ -axis, $y = p(x) = 0$ :

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

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

Step 2 · Find the Intersection with the $y$ -axis

On the $y$ -axis, $x = 0$ :

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

So, 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 Line

To graph $y = 3x + 2$ , we list the key points:

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

Plot the points $\left(-\dfrac{2}{3}, 0\right)$ , $(0, 2)$ , $(1, 5)$ , and $(3, 11)$ on the coordinate plane and join them with a straight line.

Step 2 · Verify the Solutions

From the plotted graph:

The line passes through $(1, 5)$ and $(3, 11)$ , verifying the polynomial $p(x) = 3x + 2$ .
The line intersects the $x$ -axis at $\left(-\dfrac{2}{3}, 0\right)$ and the $y$ -axis at $(0, 2)$ , verifying the intercept coordinates.

Answer

(iii) The straight line graph of $y = 3x + 2$ passes through $(1, 5)$ and $(3, 11)$ and intersects the axes at $\left(-\dfrac{2}{3}, 0\right)$ and $(0, 2)$ , verifying all calculated values.

Common Mistakes

Swapping Axis Conditions: Confusing $x = 0$ for the $x$ -intercept and $y = 0$ for the $y$ -intercept. Always remember that the $x$ -axis has $y = 0$ and the $y$ -axis has $x = 0$ .
Sign Error in Intercept: Solving $3x + 2 = 0$ as $x = \dfrac{2}{3}$ instead of $x = -\dfrac{2}{3}$ .

Question 10

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