Question

How do you graph $y=\dfrac{1}{4}x-4$ by plotting points? \[\]

Answer 1

Hint: We recall the three forms of writing a linear equation, the general form$Ax+By+C=0$, the slope intercept form $y=mx+c$ .We recall that we need points to plot a line. We find the first point as the intercepted point at $x=0$ and we take any other value of $x$ to get the other point. \[\]

Complete step-by-step solution:
We know from the Cartesian coordinate system that every linear equation $Ax+By+C=0$can be represented as a line. If the line is inclined with positive $x-$axis at an angle $\theta $ then its slope is given by $m=\tan \theta $ and if it cuts $y-$axis at a point $\left( 0,c \right)$ from the origin the $y-$intercept is given by $c$. The slope-intercept form of equation is given by
\[y=mx+c\]
We know that if $m$ is positive then we get increasing from left to right and if $m$ is negative we get a line decreasing from left to right. We are given in the following equation in the question,
\[y=\dfrac{1}{4}x-4\]
We see that the above equation is in slope-intercept form with slope $m=\dfrac{1}{4}>0$ and $y-$intercept $c=-4$ . So the given line increases from left to right and passes through the intercepts $y-$axis at $\left( 0,c \right)=\left( 0,-4 \right)$. We need another point to draw the line. Let us take $x=4$ and have;
\[y=\dfrac{1}{4}\cdot 4-4=1-4=-3\]
So the other point is $\left( 4,-3 \right)$. We plot $\left( 0,-4 \right),\left( 4,-3 \right)$ and join them with a line. \[\]

Note: We note to take $x$ as multiple of 4 to find the second point for integral value of $y$.We can also take $x-$intercept $y=0\Rightarrow x=16$ to plot the other point. If the slope $m=0$ we get a line parallel to the $x-$axis and if the slope is undefined which means $m=\infty $then we get a line perpendicular to $x-$axis.