Question

8
creating/writing equations:
part a: create an equation with a slope of \\(\frac{2}{3}\\) and a y-intercept of \\(-4\\).
this will be a \\(\boldsymbol{\vee}\\) function. the equation is \\(y = \boldsymbol{}\\).

part b: create an equation based on the table below.
\\(\

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

\\)
the table represents a \\(\boldsymbol{\vee}\\) function. the equation is \\(y = \boldsymbol{}\\)

Explanation:

Part A:

Step 1: Recall the slope - intercept form

The slope - intercept form of a linear function is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept.

Step 2: Substitute the given values

We are given that the slope $m=\frac{2}{3}$ and the y - intercept $b = - 4$. Substituting these values into the slope - intercept form, we get $y=\frac{2}{3}x-4$. Since it is a linear equation, it is a linear function.

Part B:

Step 1: Check if it is a function

A relation is a function if for every input $x$ (independent variable) there is exactly one output $y$ (dependent variable). Looking at the table, for $x = 0$, $y=-2$; for $x = 1$, $y = 1$; for $x=2$, $y = 2$; for $x = 3$, $y=1$. Each $x$ has exactly one $y$, so it is a function.

Step 2: Analyze the type of function

First, let's check if it is linear. The slope between $(0,-2)$ and $(1,1)$ is $m_1=\frac{1 - (-2)}{1-0}=\frac{3}{1}=3$. The slope between $(1,1)$ and $(2,2)$ is $m_2=\frac{2 - 1}{2 - 1}=\frac{1}{1}=1$. Since the slopes are not equal, it is not a linear function. Let's assume it is a quadratic function. The general form of a quadratic function is $y=ax^{2}+bx + c$.
When $x = 0$, $y=-2$, so $c=-2$.
When $x = 1$, $y = 1$, so $a(1)^{2}+b(1)-2=1\Rightarrow a + b=3$.
When $x = 2$, $y = 2$, so $a(2)^{2}+b(2)-2=2\Rightarrow4a + 2b=4\Rightarrow2a + b = 2$.
Subtract the first equation $a + b=3$ from $2a + b=2$:
$(2a + b)-(a + b)=2 - 3\Rightarrow2a + b - a - b=-1\Rightarrow a=-1$.
If $a=-1$, then from $a + b=3$, we have $-1 + b=3\Rightarrow b = 4$.
So the quadratic function is $y=-x^{2}+4x - 2$. Let's check for $x = 3$: $y=-9 + 12-2=1$, which matches the table.

Answer:

Part A:

This will be a linear function. The equation is $y=\frac{2}{3}x - 4$.

Part B:

The table represents a quadratic function. The equation is $y=-x^{2}+4x - 2$.