find the domain & range of each relation & identify any relations that define functions
y=3(x+1)^2-5
To find the domain and range of the given relation y = 3(x+1)^2 - 5, we need to understand the properties of the equation.
First, let's analyze the domain. The domain represents all possible values of x for which the equation is defined or valid.
Since the equation is a quadratic equation with no restrictions or limitations, the domain can be any real number. Therefore, the domain of this relation is (-∞, ∞) or all real numbers.
Next, let's determine the range. The range represents all possible values that y can take based on the given equation.
In this quadratic equation, we have a quadratic term (x+1)^2, which is always greater than or equal to zero. This means that the y-values (output) will have a minimum value but no maximum value.
By observing the equation y = 3(x+1)^2 - 5, we can see that the term 3(x+1)^2 will always be positive or zero, as it is squared. So, the smallest value y can take is -5.
Hence, the range of this relation is (-5, ∞) or all real numbers greater than -5.
Now, let's determine if this relation defines a function. A relation defines a function if no x-value is paired with multiple y-values. In other words, each x-value should have a unique y-value.
In the given equation, y = 3(x+1)^2 - 5, for every x-value, there is only one corresponding y-value. Therefore, this relation does define a function.
To summarize:
Domain: (-∞, ∞) or all real numbers
Range: (-5, ∞) or all real numbers greater than -5
Defines a function: Yes