/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Find the gradient, \(m\), and \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 1

Find the gradient, \(m\), and \(y\)-intercept, \(c\), of the following graphs, and sketch them: a \(y=x+1\) b \(y=x-1\) c \(y=3 x+5\) d \(y=\frac{1}{2} x-7\) e \(y-\pi=0\) f \(y+\pi-x=0\)

Short Answer

Expert verified
a) m=1, c=1; b) m=1, c=-1; c) m=3, c=5; d) m=1/2, c=-7; e) m=0, c=蟺; f) m=1, c=-蟺

Step by step solution

01

- Understand the Slope-Intercept Form

The slope-intercept form of a linear equation is given by: \[ y = mx + c \]where \(m\) is the gradient (or slope) and \(c\) is the y-intercept.
02

- Analyze Equation a

The equation for part (a) is: \[ y = x + 1 \]Here, \(m = 1\) and \(c = 1\). Therefore, the gradient, \(m\), is 1 and the y-intercept, \(c\), is 1.
03

- Analyze Equation b

The equation for part (b) is: \[ y = x - 1 \]Here, \(m = 1\) and \(c = -1\). Therefore, the gradient, \(m\), is 1 and the y-intercept, \(c\), is -1.
04

- Analyze Equation c

The equation for part (c) is: \[ y = 3x + 5 \]Here, \(m = 3\) and \(c = 5\). Therefore, the gradient, \(m\), is 3 and the y-intercept, \(c\), is 5.
05

- Analyze Equation d

The equation for part (d) is: \[ y = \frac{1}{2}x - 7 \]Here, \(m = \frac{1}{2}\) and \(c = -7\). Therefore, the gradient, \(m\), is \( \frac{1}{2} \) and the y-intercept, \(c\), is -7.
06

- Analyze Equation e

The equation for part (e) is: \[ y - \pi = 0 \]Simplify to get: \[ y = \pi \]Here, there is no \(x\) term, so \(m = 0\) and \(c = \pi\). Therefore, the gradient, \(m\), is 0 and the y-intercept, \(c\), is \(\pi\).
07

- Analyze Equation f

The equation for part (f) is: \[ y + \pi - x = 0 \]Simplify to get: \[ y = x - \pi \]Here, \(m = 1\) and \(c = -\pi\). Therefore, the gradient, \(m\), is 1 and the y-intercept, \(c\), is -\pi.
08

- Sketch the Graphs

Using the gradients and y-intercepts found, draw each line on a coordinate plane. The slope (gradient) determines the angle of the line, and the y-intercept shows where the line crosses the y-axis.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope-Intercept Form
Understanding the slope-intercept form is essential for identifying the gradient and the y-intercept in linear equations. The general format is given by: \[ y = mx + c \] Here, \(m\) represents the gradient (or slope) of the line, and \(c\) is the y-intercept where the line crosses the y-axis. When a linear equation is arranged in this form, it is easy to read off both the gradient and y-intercept values directly. For example, in the equation \(y = 2x + 3\), the gradient \(m\) is 2, and the y-intercept \(c\) is 3. Recognizing this format allows you to quickly interpret the characteristics of a linear graph.
Gradient
The gradient, also known as the slope, indicates how steep a line is. It is represented by the symbol \(m\) in the slope-intercept form equation \(y = mx + c\). The gradient can be positive, negative, zero, or undefined:
  • A positive gradient means the line slopes upwards as you move from left to right.
  • A negative gradient means the line slopes downwards as you move from left to right.
  • A zero gradient means the line is horizontal.
  • An undefined gradient means the line is vertical.
For example, the equation \(y = 3x + 5\) has a gradient \(m = 3\), indicating a steep upward slope. Conversely, the equation \(y = -2x + 7\) has a gradient \(m = -2\), suggesting a downward slope. Identifying the gradient helps in understanding the direction and steepness of the line.
Y-Intercept
The y-intercept is where the line crosses the y-axis on a graph. It is represented by the symbol \(c\) in the slope-intercept form equation \(y = mx + c\). This value shows where the line intersects the y-axis when the value of \(x\) is zero. For example, in the equation \(y = x + 1\), the y-intercept \(c\) is 1. This means that the line crosses the y-axis at \(y = 1\). If we consider another equation \(y = 3x - 2\), the y-intercept \(c\) is -2, indicating the line crosses the y-axis at \(y = -2\). Knowing the y-intercept allows you to pinpoint the exact location where the line meets the y-axis, which is crucial for sketching the graph accurately.
Graphing Lines
Graphing lines requires a clear understanding of both the gradient and y-intercept. First, start by plotting the y-intercept on the coordinate plane. This is the point where the line crosses the y-axis. Next, use the gradient to determine the direction and slope of the line. For a gradient \(m\), this means starting from the y-intercept and moving up or down by \(m\) units while moving one unit horizontally. For instance, a line with gradient \(m = 2\) means moving up 2 units for every 1 unit moved to the right. Taking some examples from the initial exercise:
  • For the equation \(y = x + 1\), the y-intercept is 1, and the gradient is 1. Start at (0, 1) and move up 1 unit for every 1 unit to the right.
  • For \(y = 3x + 5\), start at (0,5) and move up 3 units for every 1 unit to the right.
Drawing these lines on the graph helps visualize the relationship between the equations and their corresponding lines, aiding in better understanding linear equations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

[electrical principles] The voltage, \(V\), across a variable load resistor of resistance \(R_{\mathrm{L}}\), is given by $$ V=\frac{E R_{\mathrm{L}}}{R+R_{\mathrm{L}}} $$ where \(E\) is the source e.m.f. and \(R\) is the source resistance. Plot the graphs (on different axes) of \(V\) versus \(R_{\mathrm{L}}\) for the corresponding values: a \(E=60\) volts, \(R=10 \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 20\) b \(E=15\) volts, \(R=3 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 6 \times 10^{3}\) c \(E=10\) volts, \(R=15 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 30 \times 10^{3}\) From each of your graphs, determine the value of \(V\) at \(R_{\mathrm{L}}=R\). Do you notice any relationship between \(V\) and \(E\) at \(R_{\mathrm{L}}=R\) ? Show algebraically that if \(R_{\mathrm{L}}=R\) then \(V=\frac{E}{2}\).

across a capacitor is given by $$ V= \begin{cases}2 t & 0 \leq t<1 \\ 4-2 t & t \geq 1\end{cases} $$ Sketch the graph of \(V\) against \(t\) for \(0 \leq t \leq 2\) [Hint: The graph of \(V\) is given in two parts. You sketch \(V=2 t\) for \(0 \leq t<1\) and then sketch \(V=4-2 t\) for \(t \geq 1 .]\)

[mechanics] The velocity, \(v\), of a particle is given by $$ v=48-3 t^{2} \quad \text { where } 0 \leq t \leq 4 $$ Sketch the graph of \(v\) against \(t\). What can you conclude about the acceleration of the particle?

Sketch the following graphs: a \(y=3\) b \(x=3\) c \(2 x+2 y=6\) d \(x+2 y=-3\) e \(x-y-3=0\) f \(2 x+7 y-5=0\)

[mechanics] The displacement, s, of an object is given by \(s=t^{3}-2 t^{2}+5\). Plot the graph of \(s\) against \(t\) for \(t\) between 0 and 3 .
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.