The general form of a quadratic function is
y = ax2 + bx + c
Domain is all real values of x for which the given quadratic function is defined.
Range is all real values of y for the given domain (real values values of x).
The general form a quadratic function is
y = ax2 + bx + c
The domain of any quadratic function in the above form is all real values.
Because, in the above quadratic function, y is defined for all real values of x.
Therefore, the domain of the quadratic function in the form y = ax2 + bx + c is all real values.
That is,
Domain = {x | x ∈ R}
To know the range of a quadratic function in the form
y = ax2 + bx + c,
we have to know the following two stuff.
They are,
(i) Parabola is open upward or downward
(ii) y-coordinate at the vertex of the Parabola .
Let us see, how to know whether the graph (parabola) of the quadratic function is open upward or downward.
(i) Parabola is open upward or downward :
y = ax2 + bx + c
If the leading coefficient or the sign of "a" is positive, the parabola is open upward and "a" is negative, the parabola is open downward.
(ii) y-coordinate at the vertex :
To know y - coordinate of the vertex, first we have to find the value "x" using the formula given below.
x = -b/2a
Now, we have to plug x = -b/2a in the given quadratic function.
So, y - coordinate of the quadratic function is
y = f(-b/2a)
How to find range from the above two stuff :
(i) If the parabola is open upward, the range is all the real values greater than or equal to
y = f(-b/2a)
(i) If the parabola is open downward, the range is all the real values less than or equal to
y = f(-b/2a)
Problem 1 :
Find the domain and range of the quadratic function given below.
y = x2 + 5x + 6
Solution :
Domain :
In the quadratic function, y = x2 + 5x + 6, we can plug any real value for x.
Because, y is defined for all real values of x.
Therefore, the domain of the given quadratic function is all real values.
That is,
Domain = {x | x ∈ R}
Range :
Comparing the given quadratic function y = x2 + 5x + 6 with
y = ax2 + bx + c
we get
a = 1
b = 5
c = 6
Since the leading coefficient "a" is positive, the parabola is open upward.
Find the x-coordinate at the vertex.
x = -b / 2a
Substitute 1 for a and 5 for b.
x = -5/2(1)
x = -5/2
x = -2.5
Substitute -2.5 for x in the given quadratic function to find y-coordinate at the vertex.
y = (-2.5)2 + 5(-2.5) + 6
y = 6.25 - 12.5 + 6
y = - 0.25
So, y-coordinate of the vertex is -0.25
Because the parabola is open upward, range is all the real values greater than or equal to -0.25
Range = {y | y ≥ -0.25}
To have better understanding on domain and range of a quadratic function, let us look at the graph of the quadratic function y = x2 + 5x + 6.
When we look at the graph, it is clear that x (Domain) can take any real value and y (Range) can take all real values greater than or equal to -0.25
Problem 2 :
Find the domain and range of the quadratic function given below.
y = -2x2 + 5x - 7
Solution :
Domain :
In the quadratic function, y = -2x2 + 5x - 7, we can plug any real value for x.
Because, y is defined for all real values of x
Therefore, the domain of the given quadratic function is all real values.
That is,
Domain = {x | x ∈ R}
Range :
Comparing the given quadratic function y = -2x2 + 5x - 7 with
y = ax2 + bx + c
we get
a = -2
b = 5
c = -7
Since the leading coefficient "a" is negative, the parabola is open downward.
x = -b / 2a
Substitute -2 for a and 5 for b.
x = -5/2(-2)
x = -5/(-4)
x = 5/4
x = 1.25
Substitute 1.25 for x in the given quadratic function to find y-coordinate at the vertex.
y = -2(1.25)2 + 5(1.25) - 7
y = -3.125 + 6.25 - 7
y = -3.875
So, y-coordinate of the vertex is -3.875.
Because the parabola is open downward, range is all the real values greater than or equal to -3.875.
Range = {y | y ≤ -3.875}
To have better understanding on domain and range of a quadratic function, let us look at the graph of the quadratic function y = -2x2 + 5x - 7.
When we look at the graph, it is clear that x (Domain) can take any real value and y (Range) can take all real values less than or equal to -3.875
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