Question Video: Determining the Sign of Quadratic Functions | Nagwa Question Video: Determining the Sign of Quadratic Functions | Nagwa

Question Video: Determining the Sign of Quadratic Functions Mathematics • First Year of Secondary School

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Determine the sign of the function 𝑓(π‘₯) = (2π‘₯ βˆ’ 1)Β².

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Video Transcript

Determine the sign of the function 𝑓 of π‘₯ equals two π‘₯ minus one squared.

The sign of a function is a description that tells us whether the function is positive or negative, or possibly even zero. We say that the function is positive when 𝑓 of π‘₯ is greater than zero, and it’s negative for values of π‘₯, where 𝑓 of π‘₯ is less than zero. And of course, we can actually use the graph to help us determine the sign of the function. 𝑓 of π‘₯ is, of course, the output for a given input of π‘₯. So, 𝑓 of π‘₯ is said to be positive when the graph of the function lies above the π‘₯-axis and it’s said to be negative when the graph of 𝑓 of π‘₯ lies below the π‘₯-axis.

So, let’s begin then by sketching the graph of 𝑓 of π‘₯ equals two π‘₯ minus one squared. If we were to distribute the parentheses here, we’d notice that we have a quadratic equation. Specifically, the coefficient of π‘₯ squared, if we were to distribute these parentheses, would be four; it’s positive. And so, we know that not only is it a parabola, it’s a U-shaped parabola since the coefficient of π‘₯ squared is positive.

Next, we can find the values of the π‘₯-intercepts of the graph of our function by letting 𝑓 of π‘₯ equal zero. In doing so, we get zero equals two π‘₯ minus one squared. We take the square root of both sides to get zero equals two π‘₯ minus one. And then if we add one to both sides, our equation is two π‘₯ equals one. Finally, we divide by two. When we do, we find that π‘₯ equals one-half is a solution to the equation 𝑓 of π‘₯ equals zero. This means there is a single π‘₯-intercept, one-half.

So, the graph of the function 𝑦 equals 𝑓 of π‘₯ or 𝑦 equals two π‘₯ minus one squared will look like this. The function is equal to zero at π‘₯ equals one-half. However, for all other values of π‘₯, we notice that the graph of the function lies above the π‘₯-axis. Therefore, 𝑓 of π‘₯ is positive for all other values of π‘₯, for all real values of π‘₯ not including π‘₯ equals one-half.

We can therefore say that the sign of the function is positive when π‘₯ is any real number not including one-half and it’s equal to zero when π‘₯ equals one-half.

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