How do you find the domain and range and determine whether the relation is a function given :#x=2y^2-3#?

Question

Genesis Allen · Accepted Answer

See below.

For y as a function of x.

#x = 2y^2 -3#

Rearrange;

Add 3 to both sides:

#2y^2 = x + 3#

Divide by 2:

#y^2 = 1/2x + 3/2 => y^2 = (x + 3)/2#

Taking roots of both sides:

#y = sqrt((x + 3)/2)#

For positive root:

For the domain we must make sure the numerator is #>= 0#, since we are only interested in real numbers.

The numerator is #>=0# for #x >= -3#

So the domain is #{ x in RR | x >= -3 }#

The range for the positive root is:

As
#x -> +oo#
#y -> +oo#

So the range is:

#{y in RR | y >= 0 }#

For negative root the domain is the same for real numbers. The range is in the reverse direction:

#{ y in RR | y <= 0 }#

The determination of a function.

For positive root only a function is defined.
For negative root only a function is defined.

For #+-sqrt((3 + x )/2 )#

This is not a function. A function is only defined if an element in a domain is mapped onto one and one only element in a range. In the case of both roots this would map one element in the domain to two elements in the range. This is sometimes called a one to many relationship.

See graph:

Jameson Allen · Answer

undefined To find the domain and range of the relation x = 2y^2 - 3 and determine if it is a function:

1. **Domain:** The domain of the relation is all possible values of y that will produce a real number for x. In this case, there are no restrictions on y, so the domain is all real numbers.

2. **Range:** To find the range, we need to determine all possible values of x that the relation can produce. Since the equation is quadratic, it opens upwards (positive coefficient of y^2), and the lowest point occurs when y = 0. Substituting y = 0 into the equation, we get x = 2(0)^2 - 3 = -3. Therefore, the range is all real numbers greater than or equal to -3.

3. **Function:** To determine if the relation is a function, we need to check if each input value of y corresponds to exactly one output value of x. Since the equation is quadratic, for each value of y, there are two possible values of x (one positive and one negative), so the relation is not a function.

Nathan Axel · Answer

undefined To find the domain and range and determine whether the relation $x = 2y^2 - 3$ is a function:

1. **Domain**: The domain of the relation refers to all possible values of $y$ for which $x$ is defined. Since $x$ is defined for all real numbers, the domain is all real numbers.

2. **Range**: The range of the relation refers to all possible values of $x$ resulting from the chosen values of $y$. To find the range, we need to solve the equation for $y$ in terms of $x$:

$$ x = 2y^2 - 3 $$

$$ x + 3 = 2y^2 $$

$$ \frac{x + 3}{2} = y^2 $$

$$ y = \pm \sqrt{\frac{x + 3}{2}} $$

Since $y$ can take both positive and negative square roots, the range is all real numbers.

3. **Function**: To determine if the relation is a function, we must ensure that each input $y$ corresponds to exactly one output $x$. If there are multiple values of $x$ for a single value of $y$, the relation is not a function.

In this case, since $y$ can take both positive and negative square roots, for each $y$, there are two possible values of $x$. Therefore, the relation is not a function.