Math 9 Chapter 2 Lesson 2: First order functions
1. Summary of theory
1.1. Define
A firstorder function is a function given by the formula \(y = ax + b,\) where \(a, b\) are the given numbers and \(a ≠ 0.\)
For example, \(y=5x;y=2x+1\) are first order functions.
1.2. Nature
The first order function \(y = ax + b\) determines for all values of x in R and has the following property:
a) Covariable on R when \(a > 0\)
b) Inverse on R when \(a < 0.\)
2. Illustrated exercise
2.1. Basic exercises
Question 1: Determine \(m\) so that the function \(y=(m1)x+2\) is covariant
Solution guide
The given function is covariant when \(m1>0\) or \(m>1\)
Verse 2: Given the function \(y=2x^2+3\). Is this function a first order function?
Solution guide
The given function is not a first order function because there is no form \(y=ax+b\)
Question 3: Given the function \(y=ax+1\). Know the graph of the function passing through the point \(A(1;2)\). What is \(a\) equal to?
Solution guide
The graph of the function passes through the point \(A(1;2)\) so \(2=a.1+1\) or \(a=1\)
2.2. Advanced exercises
Question 1: Determine the line passing through the two points \(A\) and \(B\), knowing that \(A(2;0)\) and \(B(0,1)\)
Solution guide
Assume the line is of the form \(y=ax+b\) with a different \(a\)
\(A\) and \(B\) belong to a straight line, so we have \(0=a.(2)+b\) and \(1=a.0+b\). Solving the system we get \(a=\frac{1}{2}\) and \(b=1\). So \(y=\frac{1}{2}x+1\)
Verse 2: Prove that if a line does not pass through the origin, intersects the horizontal axis at a point with coordinates a, and intersects the vertical axis at a point with coordinates b, then the line has the equation
\(\frac{x}{a}+\frac{y}{b}=1\)
Solution guide
Assume the line is of the form \(y=mx+n\) with a different \(m\)
The line passes through the point \((0;b)\) so \(b=m.0+n=>n=b\)
The line passes through the point \((a;0)\) so \(0=m.a+b=>m=\frac{b}{a}\) (notice that \(a\) is different )
From there: \(y=\frac{b}{a}x+b\) or \(\frac{y}{b}=\frac{x}{a}+1\) ie \( \frac{x}{a}+\frac{y}{b}=1\)
3. Practice
3.1. Essay exercises
Question 1: Which of the following functions is a firstorder function? Let’s determine their coefficients \(a,\ b\) and see which firstorder functions are covariant and inverse.
a) \(y = 1 – 5x\)
b) \(y = 0.5x\)
c) \(y = \sqrt 2 \left( {x – 1} \right) + \sqrt 3 \)
d) \(y=2x^2+3\)
Verse 2: Given a first order function \(y = \left( {m + 1} \right)x + 5.\)
a) Find the value of m so that the function y is a covariate function;
b) Find the value of m so that the function y is an inverse function.
Question 3: A rectangle has dimensions \(20cm\) and \(30cm\). People reduce each size of that shape by \(x\) \((cm)\) to get a new rectangle with perimeter \(y\) \((cm)\). Make a formula to calculate \(y\) in terms of \(x\).
Question 4: For what values of \(m\) each of the following functions is a first order function?
a) \(y=\sqrt{5 – m}(x – 1)\);
b) \(y = \dfrac{m + 1}{m – 1}x +3,5\)
Question 5: Prove that the first order function y = ax + b is covariate when a > 0 and inverse when a < 0.
3.2. Multiple choice exercises
Question 1: Determine \(m\) so that the function \(y=(m3)x+1\) is inverse
A. 2
B. 4
C. 3
D. No square meter
Verse 2: For what value of \(m\) is the function \(y=\sqrt{1m}.x+1\) a first order function?
A. No square meter
B. Countless m
C. 2
D. \(m<1\)
Question 3: Given three points \(A(0;5), B(1;2), C(2;1)\). Ask what these three points form?
A. Straight line
B. Isosceles triangle
C. Right triangle
D. Pointed triangle
Question 4: Given the function \(y=ax+2\). Know the graph of the function passing through the point \(A(1;0)\). What is \(a\) equal to?
A. 2
B. 1
C. 2
D. Can’t find
Question 5: Which of the following is not a first order function?
A. \(y=x+1\)
B. \(y=x^2\)
C. \(y=2x\)
D. \(y=2\sqrt{3}x+1\)
4. Conclusion
Through this lesson, students will learn the following key topics:

Students know that the general form of the first order function is y = ax + b (a 0)

Understand the properties of firstorder functions on the deterministic set of variables, covariance and inverse.
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