# Given linear equation y = -3 a. find the y-intercept and slope. b. determine whether the line slopes upward, slopes downward, or is horizontal, without graphing the equation. c.use two points to graph the equation

Given linear equation y = -3
a. find the y-intercept and slope.
b. determine whether the line slopes upward, slopes downward, or is horizontal, without graphing the equation.
c.use two points to graph the equation
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(a) y-intercept and slope: In a linear equation $y={b}_{0}+{b}_{1}x$, the constant b1 be the slope and b0 be the y-intercept form and x is the independent variable and y is the dependent variable.
Comparing the given equation with the general form of linear equation the slope of the equation is 0 and the y-intercept is –3. Thus, the slope of the linear equation is ${b}_{1}=0$ and the y-intercept is ${b}_{0}=-3$
(b) It is known that, the slope of the linear equation $y={b}_{0}+{b}_{1}x$ is upward if ${b}_{1}>0$, the slope of the linear equation $y={b}_{0}+{b}_{1}x$ is downward if ${b}_{1}<0$, and the slope of the linear equation $y={b}_{0}+{b}_{1}x$ is horizontal if ${b}_{1}=0$ .
Thus, in the given equation $y=-3{b}_{1}=0$
Thus, the slope is horizontal.
(c) The two points $\left({x}_{1},{y}_{1}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left({x}_{2},{y}_{2}\right)$ on the given line are obtained
If $x=0$
$y=-3$
Thus, one point on the line is $\left({x}_{1},{y}_{1}\right)=\left(0,-3\right).$
If $x=1$
$y=-3$
Thus, the second point on the line is $\left({x}_{2},{y}_{2}\right)=\left(1,-3\right)$