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Introduction to Polynomials - Interpret slope and y-intercept to visualise linear relationships in y = ax + b

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A linear polynomial is a polynomial of degree 1, generally expressed as P(x)=ax+bP(x) = ax + b, where a≠0a \neq 0. In a coordinate plane, this represents a straight line y=ax+by = ax + b.

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The coefficient aa is known as the slope or gradient of the line. It measures the steepness and direction of the line. If a>0a > 0, the line moves upwards from left to right. If a<0a < 0, it moves downwards.

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The constant term bb is the y-intercept. This is the point where the line crosses the yy-axis. At this point, the value of xx is always 00, giving the coordinates (0,b)(0, b).

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The magnitude of aa determines the steepness. A larger absolute value of aa results in a steeper line, while a smaller value (closer to 00) results in a flatter line.

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The zero of the linear polynomial ax+bax + b is the xx-intercept of the graph, found by setting y=0y = 0, which gives x=−bax = -\frac{b}{a}.

📐Formulae

y=ax+by = ax + b

a=Change in yChange in x=y2−y1x2−x1a = \frac{\text{Change in } y}{\text{Change in } x} = \frac{y_2 - y_1}{x_2 - x_1}

y-intercept=(0,b)\text{y-intercept} = (0, b)

x-intercept (Zero of polynomial)=(−ba,0)\text{x-intercept (Zero of polynomial)} = \left(-\frac{b}{a}, 0\right)

💡Examples

Problem 1:

Identify the slope and y-intercept for the linear relationship y=3x−4y = 3x - 4 and describe the graph's behavior.

Solution:

From the equation y=3x−4y = 3x - 4, comparing it with y=ax+by = ax + b, we get a=3a = 3 and b=−4b = -4.

Explanation:

The slope a=3a = 3 is positive, which means the line rises as xx increases. For every 11 unit move to the right, the line moves 33 units up. The y-intercept b=−4b = -4 means the line crosses the yy-axis at the point (0,−4)(0, -4).

Problem 2:

A taxi service charges a fixed fee of ₹50₹ 50 and an additional ₹10₹ 10 per kilometer. Write this as a linear polynomial and find the cost for 1515 km.

Solution:

Let xx be the distance in km and yy be the total cost. The relationship is y=10x+50y = 10x + 50. For x=15x = 15: y=10(15)+50y = 10(15) + 50 y=150+50y = 150 + 50 y=200y = 200

Explanation:

Here, the slope a=10a = 10 represents the rate per km, and the y-intercept b=50b = 50 represents the fixed starting cost. The total cost for 1515 km is ₹200₹ 200.

Problem 3:

Calculate the slope of a line that passes through the points (2,5)(2, 5) and (4,9)(4, 9).

Solution:

Using the slope formula: a=y2−y1x2−x1a = \frac{y_2 - y_1}{x_2 - x_1} a=9−54−2a = \frac{9 - 5}{4 - 2} a=42=2a = \frac{4}{2} = 2

Explanation:

The slope is 22. This means for every unit increase in xx, the value of yy increases by 22 units.