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How do you graph \[y = - 3x + 5\] using slope intercept form?

seo-qna
Last updated date: 05th Mar 2024
Total views: 341.7k
Views today: 3.41k
IVSAT 2024
Answer
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Hint: In this question, we draw a graph using the above equation. The equation which is given above is a straight line equation. In the above equation is an independent variable and y is a dependent variable. For drawing the graph, first we calculate the points and points are calculated as we put the value of x like \[\left( {0,\;1,\;2,.......} \right)\] in the above equation and then find the value of y. then we draw the graph by using these points.

Complete step by step answer:
Now we come to question, in the question the straight-line equation is given as below.
\[ \Rightarrow y = - 3x + 5\]
The slope is \[m = - 3\].
Now we find the points from the above equation.
We put the value of \[x = 0\] in the above equation, and then we find the value of y.
\[
  y = - 3 \times 0 + 5 \\
  y = 5 \\
 \]
The value of y is \[ = 5\]
Then the point is \[\left( {0,\;5} \right)\].
Now, we put the value of \[x = 1\]in the above equation, and then find the value of y.
\[
  y = - 3 \times 1 + 5 \\
  y = 2 \\
 \]
The value of y is \[ = 2\]
Then the point is \[\left( {1,\;2} \right)\].
Then, we put \[x = 2\] in the above equation then \[y = - 1\]. Then the point is \[\left( {2,\; - 1} \right)\].
Hence, we calculate the same for other points.
Now we draw the graph by using the points.
The points are \[\left( {\left( {0,\;5} \right),\left( {1,\;2} \right),\left( {2,\; - 1} \right)} \right)\].
Therefore, the graph is drawn as below.
seo images


Note:
In this question, we use the word slope intercept form. Slope intercept form is defined as the form of straight-line equation. Slope intercept form is represented as below.
\[ \Rightarrow y = mx + c\]
Where,
\[m = \] Slope
\[c = \] It is y intercept
And x and y are the variable. Where x is an independent variable and y is a dependent variable. When we put the value of x then we find the value of y.
draw the graph.
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