How do you graph \[y=\sqrt{x-3}\]?
Answer
Verified
447.3k+ 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
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
The reservoir of dam is called Govind Sagar A Jayakwadi class 11 social science CBSE
10 examples of friction in our daily life
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
State the laws of reflection of light