How do you graph \[y=\sqrt{x-3}\]?
Answer
630.6k+ views
Hint: The given equation is not a linear equation. We can write the given equation as \[{{y}^{2}}=x-3\].so the given equation is a parabola. Here, x and y take only positive values. So, the graph lies only in the first quadrant.
Complete step by step answer:
As per the given question, we are given a function which we have to graph. And, the given function is \[y=\sqrt{x-3}\].
We know that the square root is always positive. Therefore, \[x-3\] should always be positive.
\[\Rightarrow x-3\underline{ > }0\to x\underline{ > }3\]
From the above equation, we can say that the graph lies from \[x=3\] to towards the positive x-axis.
The graph does not touch the y-axis also.
Since the right hand side of the equation is always positive, the value of y cannot be negative.
Now let us substitute some points to plot the graph. Let y=0 then
\[\Rightarrow y=\sqrt{x-3}\Rightarrow 0=\sqrt{x-3}\]
Now squaring on both sides
\[\begin{align}
& \Rightarrow y=\sqrt{x-3} \\
& \Rightarrow 0=\sqrt{x-3} \\
& \Rightarrow {{0}^{2}}={{(\sqrt{x-3})}^{2}} \\
& \Rightarrow 0=\pm (x-3) \\
& \Rightarrow x=3 \\
\end{align}\]
\[\therefore \] The point lying on the x-axis is \[(3,0)\].
Now substitute \[x=4\] in the equation.
\[\begin{align}
& \Rightarrow y=\sqrt{x-3} \\
& \Rightarrow y=\sqrt{4-3} \\
& \Rightarrow y=\pm 1 \\
\end{align}\]
Since y cannot be negative, \[y=1\].
\[\therefore \] The point on the graph is \[(4,1)\]
Let us take \[x=7\] then
\[\begin{align}
& \Rightarrow y=\sqrt{x-3} \\
& \Rightarrow y=\sqrt{7-3} \\
& \Rightarrow y=\sqrt{4} \\
& \Rightarrow y=\pm 2 \\
\end{align}\]
Since y cannot be negative, \[y=2\].
\[\therefore \] The point on the graph is \[(7,2)\].
Plotting all the three points on the graph, and joining them gives a parabola lying in the first quadrant.
Note:
In order to solve these types of problems, we need to have knowledge over how to calculate the intervals of functions. Take a greater number of points to plot so that the graph will be accurate. Take the points to be plotted on the graph wisely so that they can be easily plotted on graph like if the values are integers then they can be easily plotted. We should avoid calculation mistakes to get the correct solution.
Complete step by step answer:
As per the given question, we are given a function which we have to graph. And, the given function is \[y=\sqrt{x-3}\].
We know that the square root is always positive. Therefore, \[x-3\] should always be positive.
\[\Rightarrow x-3\underline{ > }0\to x\underline{ > }3\]
From the above equation, we can say that the graph lies from \[x=3\] to towards the positive x-axis.
The graph does not touch the y-axis also.
Since the right hand side of the equation is always positive, the value of y cannot be negative.
Now let us substitute some points to plot the graph. Let y=0 then
\[\Rightarrow y=\sqrt{x-3}\Rightarrow 0=\sqrt{x-3}\]
Now squaring on both sides
\[\begin{align}
& \Rightarrow y=\sqrt{x-3} \\
& \Rightarrow 0=\sqrt{x-3} \\
& \Rightarrow {{0}^{2}}={{(\sqrt{x-3})}^{2}} \\
& \Rightarrow 0=\pm (x-3) \\
& \Rightarrow x=3 \\
\end{align}\]
\[\therefore \] The point lying on the x-axis is \[(3,0)\].
Now substitute \[x=4\] in the equation.
\[\begin{align}
& \Rightarrow y=\sqrt{x-3} \\
& \Rightarrow y=\sqrt{4-3} \\
& \Rightarrow y=\pm 1 \\
\end{align}\]
Since y cannot be negative, \[y=1\].
\[\therefore \] The point on the graph is \[(4,1)\]
Let us take \[x=7\] then
\[\begin{align}
& \Rightarrow y=\sqrt{x-3} \\
& \Rightarrow y=\sqrt{7-3} \\
& \Rightarrow y=\sqrt{4} \\
& \Rightarrow y=\pm 2 \\
\end{align}\]
Since y cannot be negative, \[y=2\].
\[\therefore \] The point on the graph is \[(7,2)\].
Plotting all the three points on the graph, and joining them gives a parabola lying in the first quadrant.
Note:
In order to solve these types of problems, we need to have knowledge over how to calculate the intervals of functions. Take a greater number of points to plot so that the graph will be accurate. Take the points to be plotted on the graph wisely so that they can be easily plotted on graph like if the values are integers then they can be easily plotted. We should avoid calculation mistakes to get the correct solution.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

