Question Video: Determining the Domain and Range of a Quadratic Function | Nagwa Question Video: Determining the Domain and Range of a Quadratic Function | Nagwa

Question Video: Determining the Domain and Range of a Quadratic Function Mathematics

Determine the domain and the range of the function 𝑓(π‘₯) = 4(π‘₯ βˆ’ 4)Β² βˆ’ 3.

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Video Transcript

Determine the domain and the range of the function 𝑓 of π‘₯ equals four multiplied by π‘₯ minus four all squared minus three.

Firstly, we recall that the domain is the set of all values on which the function acts, which we can also think of as the set of input values to the function. As the function 𝑓 of π‘₯ is a polynomial and, more specifically, a quadratic, there are no restrictions on what values it can act on. Therefore, we say that the domain of this function is the set of all real numbers.

The range of a function is the set of all values the function produces, which we can think of as the set of all output values. To determine the range of a quadratic function, we can consider its turning point. Now, this quadratic function has been given to us in its completed square or vertex form. 𝑓 of π‘₯ equals π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž. And we know that when a quadratic function is given in this form, its vertex has the coordinates negative 𝑝, π‘ž. The value of 𝑝 for our quadratic is negative four, and the value of π‘ž is negative three. So the vertex will be at negative negative four, that’s four, negative three.

As the value of π‘Ž, the coefficient of π‘₯ squared in our quadratic function, is four, which is positive, we know that its graph will be a parabola which curves upwards. So this vertex of four, negative three will be a minimum point. The possible values of 𝑓 of π‘₯ then will be all the values from this minimum value of the function negative three upwards.

We can express this either as 𝑓 of π‘₯ is greater than or equal to negative three or using interval notation as the interval from negative three to ∞, which is closed at the lower end and open at the upper end. We can answer the problem then by saying that the domain of this function is the set of all real numbers and the range is the interval from negative three to ∞, which is closed at the lower end and open at the upper end.

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