Question

How do you graph $y={\displaystyle \frac{3}{2}}x+3$ ?

Answer 1

Hint: Change of form of the given equation will give the x-intercept and y-intercept of the line $y={\displaystyle \frac{3}{2}}x+3$. We change it to the form of ${\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}=1$ to find the x-intercept, and y-intercept of the line as $p$ and $q$ respectively. then we place the points on the axes and from there we draw the line on the graph.

Complete step by step answer:
We are taking the general equation of a line to understand the slope and the intercept form of the line $y={\displaystyle \frac{3}{2}}x+3$. The given equation is in the form of $y=mx+k$. m is the slope of the line. The slope of the line is ${\displaystyle \frac{3}{2}}$ .
We have to find the x-intercept, and y-intercept of the line $y={\displaystyle \frac{3}{2}}x+3$.
For this we convert the given equation into the form of ${\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}=1$ . From the form we get that the x intercept, and y intercept of the line will be $p$ and $q$ respectively. The points will be $(p,0),(0,q)$ .
The given equation is $y={\displaystyle \frac{3}{2}}x+3$ . Converting into the form of ${\displaystyle \frac{x}{p}}+{\displaystyle \frac{y}{q}}=1$ , we get
 $\begin{array}{rl}& y={\displaystyle \frac{3}{2}}x+3\\ & \Rightarrow y-{\displaystyle \frac{3}{2}}x=3\\ & \Rightarrow {\displaystyle \frac{x}{-2}}+{\displaystyle \frac{y}{3}}=1\end{array}$
Therefore, the x intercept, and y intercept of the line $2x+y=5$ is 2 and 3 respectively. The axes intercepting points are $(-2,0),(0,3)$ .

Note:
A line parallel to the X-axis does not intersect the X-axis at any finite distance. Hence, we cannot get any finite x-intercept of such a line. The same goes for lines parallel to the Y-axis. In the case of the slope of a line, the range of the slope is 0 to $\mathrm{\infty }$.