How do you find the domain and range for #y = x^25#?
the domain is
Next, you locate x:
that has two answers:
The graphic also displays the domain and range:
graph{x^2–5 [–10,–10,–5]}
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To find the domain and range of the function ( y = x^2  5 ):
Domain: The domain of the function represents all possible input values (xvalues) that the function can accept. Since ( y = x^2  5 ) is a polynomial function, it is defined for all real numbers.
Therefore, the domain is all real numbers, which can be expressed using interval notation as ( (\infty, \infty) ).
Range: The range of the function represents all possible output values (yvalues) that the function can produce. Since ( y = x^2  5 ) is a quadratic function, its graph is a parabola that opens upwards.
The minimum value of ( y ) occurs at the vertex of the parabola. Since the coefficient of ( x^2 ) is positive, the parabola opens upwards, and the minimum value of ( y ) is achieved when ( x = 0 ). When ( x = 0 ), ( y = 0^2  5 = 5 ).
So, the minimum value of ( y ) is 5.
Therefore, the range is all real numbers greater than or equal to 5, which can be expressed using interval notation as ( [5, \infty) ).
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To find the domain and range for the function ( y = x^2  5 ), follow these steps:

Domain: The domain of a function represents all possible values of ( x ) for which the function is defined. Since ( x ) can take any real value, the domain of ( y = x^2  5 ) is all real numbers, denoted as ( (\infty, \infty) ).

Range: The range of a function represents all possible values of ( y ) that the function can output. To find the range, analyze the behavior of the function ( y = x^2  5 ). Since the function is a quadratic function, its graph is a parabola that opens upwards. The vertex of the parabola is the lowest point, and the ycoordinate of the vertex gives the minimum value of ( y ).
To find the vertex, use the formula: [ x = \frac{b}{2a} ] where ( a ) is the coefficient of ( x^2 ) and ( b ) is the coefficient of ( x ). For the function ( y = x^2  5 ), ( a = 1 ) and ( b = 0 ), so ( x = \frac{0}{2(1)} = 0 ).
Substitute ( x = 0 ) into the function to find the ycoordinate of the vertex: [ y = (0)^2  5 = 5 ]
Therefore, the vertex of the parabola is ( (0, 5) ). Since the parabola opens upwards, the range of ( y = x^2  5 ) is all real numbers greater than or equal to the ycoordinate of the vertex, so the range is ( [5, \infty) ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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