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How do you graph the function \[f\left( x \right)=-\dfrac{1}{2}x+3\]?

Last updated date: 20th Jun 2024
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Hint: To find the graph for the given equation we have to convert the given straight line equation into intercept form. For converting our equation into intercept form we have to do some operations. After that we will get an X and Y intercept for the given equation. So, by plotting them we will get the graph

Complete step by step answer:
For the given question we have to draw a graph for the function \[f\left( x \right)=-\dfrac{1}{2}x+3\].
Let us assume the above equation as y and mark it as equation (1).
\[y=-\dfrac{1}{2}x+3.............\left( 1 \right)\]
If we convert the equation (1) to intercept form we will get two points which are used to plot the graph easily.
As we know the intercept form of a straight line equation is \[\dfrac{x}{a}+\dfrac{y}{b}=1\].
Let us consider the formula \[\dfrac{x}{a}+\dfrac{y}{b}=1\] as formula (f1).
  & \dfrac{x}{a}+\dfrac{y}{b}=1.........\left( f1 \right) \\
 & \\
Now let us convert our straight line equation into intercept form.
So, for converting equation (1) as intercept form we have to do some operations.
By transferring \[-\dfrac{1}{2}x\] from RHS (right hand side) to LHS (left hand side), we get
\[\Rightarrow y+\dfrac{1}{2}x=3\]
Let us consider
\[y+\dfrac{1}{2}x=3.......\left( 2 \right)\]
By observing equation (1) we can see that RHS of equation (1) is equal to 1.
So, let us make RHS of equation (2) to 1, for that we have to divide both sides of equation with 3.
By dividing both sides of equation (2) with 3, we get
\[\Rightarrow \dfrac{x}{6}+\dfrac{y}{3}=1\]
Let us consider
\[\dfrac{x}{6}+\dfrac{y}{3}=1..........\left( 3 \right)\]
Therefore from the equation (3), X and Y intercepts are (-6, 0) and (0, 3) respectively.
Plotting the graph using the X and Y intercept.
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Another way to do this is find any two distinct points on the graph then draw a straight line through them. For that substitute the x values in f(x). Therefore we will get ordered pairs in the form (x, f(x)). By plotting these points we will get the resultant graph.