How do you find the domain and range for #y = x^2-5#?

Answer 1

the domain is #RR#, the range is #[-5, +oo)#

Since y is a polynomial, its domain is #RR#

Next, you locate x:

#x^2=y+5#

that has two answers:

#x=+-sqrt(y+5)#
that are verified only if #y+5>=0#
or #y>=-5#
Then the range of y is #[-5, +oo)#

The graphic also displays the domain and range:

graph{x^2–5 [–10,–10,–5]}

the domain, on x-axis, is all along the line, the range begin in the y-coordinate -5 and continues to #+oo#
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Answer 2

To find the domain and range of the function ( y = x^2 - 5 ):

Domain: The domain of the function represents all possible input values (x-values) 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 (y-values) 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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Answer 3

To find the domain and range for the function ( y = x^2 - 5 ), follow these steps:

  1. 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) ).

  2. 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 y-coordinate 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 y-coordinate 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 y-coordinate of the vertex, so the range is ( [-5, \infty) ).

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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