how to find the domain of a quadratic function
DOMAIN AND RANGE OF QUADRATIC FUNCTION WORKSHEET
Trouble ane :
Find the domain and range of the quadratic function given below.
y = 10ii + 5x + 6
Problem two :
Detect the domain and range of the quadratic office given below.
y = -2x2 + 5x - 7
Solutions
Problem 1 :
Notice the domain and range of the quadratic office given below.
y = ten2 + 5x + 6
Solution :
Domain :
In the quadratic function, y = 10ii + 5x + half dozen , we can plug any existent value for ten.
Considering, y is defined for all real values of x.
Therefore, the domain of the given quadratic function is all existent values.
That is,
Domain = {x | x ∈ R}
Range :
Comparing the given quadratic role y = x2 + 5x + 6 with
y = axii + bx + c
we get
a = 1
b = 5
c = 6
Since the leading coefficient "a" is positive, the parabola is open upward.
Detect the x-coordinate at the vertecx.
10 = -b / 2a
Substitute one for a and 5 for b.
x = -5/two(ane)
x = -5/two
x = -2.5
Substitute -2.5 for x in the given quadratic function to find y-coordinate at the vertex.
y = (-two.v)2 + 5(-ii.v) + six
y = 6.25 - 12.5 + vi
y = - 0.25
So, y-coordinate of the vertex is -0.25
Because the parabola is open up, range is all the existent values greater than or equal to -0.25
Range = {y | y≥ -0.25}
To accept amend understanding on domain and range of a quadratic part, let us look at the graph of the quadratic part y = xii + 5x + 6.
When we look at the graph, information technology is articulate that x (Domain) tin can take whatsoever real value and y (Range) tin 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 whatever real value for 10.
Because, y is divers for all real values of 10
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 = axtwo + bx + c
nosotros get
a = -2
b = 5
c = -seven
Since the leading coefficient "a" is negative, the parabola is open downward.
ten = -b / 2a
Substitute -2 for a and 5 for b.
x = -5/2(-2)
x = -5/(-4)
x = v/4
x = 1.25
Substitute 1.25 for x in the given quadratic function to notice y-coordinate at the vertex.
y = -2 (i.25)2 + 5(1.25) - 7
y = -3.125 + vi.25 - seven
y = -3.875
And 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 ≤ -iii.875}
To have better understanding on domain and range of a quadratic function, let u.s.a. look at the graph of the quadratic role y = -2x2 + 5x - 7 .
When we await at the graph, information technology is articulate that ten (Domain) can take any existent value and y (Range) can take all existent values less than or equal to -3.875
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