Question

find the error dan claims that because ( y = x + 1 ) and ( y = 3x + 2 ) are both linear functions, the function ( y = (x + 1)(3x + 2) ) must also be linear. is he correct? explain your reasoning. you can choose points on the graph of the function ( y = (x + 1)(3x + 2) ) and show on a coordinate plane that they on a single line.

Explanation:

Step1: Expand the quadratic function

To determine the type of function \( y=(x + 1)(3x + 2) \), we first expand it. Using the distributive property (FOIL method), we get:

$$
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$$

Step2: Identify the function type

A linear function has the form \( y = mx + b \), where the highest power of \( x \) is 1. The expanded form of \( y=(x + 1)(3x + 2) \) is \( y = 3x^{2}+5x + 2 \), which is a quadratic function (since the highest power of \( x \) is 2). So, the product of two linear functions is a quadratic (non - linear) function. Thus, Dan is incorrect.

Step3: Choose points to verify (optional but helpful)

Let's choose two points for \( y=(x + 1)(3x + 2) \).

• When \( x = 0 \): \( y=(0 + 1)(0 + 2)=2 \), so the point is \( (0,2) \).
• When \( x = 1 \): \( y=(1 + 1)(3\times1+2)=2\times5 = 10 \), so the point is \( (1,10) \).

Now, let's find the slope between these two points. The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{10 - 2}{1 - 0}=8 \).

Now, let's take another point. When \( x=- 1 \): \( y=(-1 + 1)(3\times(-1)+2)=0\times(-1)=0 \), the point is \( (-1,0) \).

The slope between \( (-1,0) \) and \( (0,2) \) is \( \frac{2-0}{0 - (-1)} = 2 \). Since the slope is not constant, the points do not lie on a straight line, confirming it's not a linear function.

Answer:

Dan is not correct. The function \( y=(x + 1)(3x + 2) \) expands to \( y = 3x^{2}+5x + 2 \), which is a quadratic (non - linear) function, not a linear function. The product of two linear functions is a quadratic function, and the points on its graph do not lie on a single straight line (e.g., points \((-1,0)\), \((0,2)\), and \((1,10)\) have different slopes between them).