Question

1. graph $y = -2x + 1$

a. what is the $y$-intercept?
b. what is the $x$-intercept?
c. is the point $(1,3)$ a solution to the equation? defend your answer.

Explanation:

Part a

Step1: Recall slope - intercept form

The equation of a line in slope - intercept form is \(y = mx + b\), where \(m\) is the slope and \(b\) is the \(y\) - intercept.
For the equation \(y=-2x + 1\), comparing it with \(y = mx + b\), we can see that \(b = 1\).
The \(y\) - intercept is the point where \(x = 0\). When \(x = 0\), \(y=-2(0)+1=1\). So the \(y\) - intercept is the point \((0,1)\).

Step1: Recall \(x\) - intercept definition

The \(x\) - intercept is the point where \(y = 0\). So we set \(y = 0\) in the equation \(y=-2x + 1\).

Step2: Solve for \(x\)

Set \(y = 0\):
\(0=-2x + 1\)
Add \(2x\) to both sides of the equation: \(2x=1\)
Divide both sides by \(2\): \(x=\frac{1}{2}=0.5\)
So the \(x\) - intercept is the point \((\frac{1}{2},0)\) (or \((0.5,0)\)).

Step1: Substitute \(x = 1\) and \(y = 3\) into the equation

We have the equation \(y=-2x + 1\). Substitute \(x = 1\) and \(y = 3\) into the left - hand side and right - hand side of the equation.
Left - hand side (LHS): \(y = 3\)
Right - hand side (RHS): \(-2x+1=-2(1)+1=-2 + 1=-1\)

Step2: Compare LHS and RHS

Since \(LHS = 3\) and \(RHS=-1\), and \(3
eq - 1\), the point \((1,3)\) does not satisfy the equation \(y=-2x + 1\).

Answer:

The \(y\) - intercept is \((0,1)\)

Part b