Question Video: Finding the Solution Sets of Quadratic Equations from Their Graphs | Nagwa Question Video: Finding the Solution Sets of Quadratic Equations from Their Graphs | Nagwa

Question Video: Finding the Solution Sets of Quadratic Equations from Their Graphs Mathematics • Third Year of Preparatory School

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The graph shows the curve with equation 𝑦 = 𝑓(π‘₯). What is the solution set of the equation 𝑓(π‘₯) = 0?

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Video Transcript

The graph shows the curve with equation 𝑦 equals 𝑓 of π‘₯. What is the solution set of the equation 𝑓 of π‘₯ equal to zero?

First, we examine the graph and recognize the curve as a parabola. A parabola is the general shape of all quadratic graphs. A parabola is a smooth, symmetrical curve, which curves upward or, like in our case, curves downward. The orientation of the parabola depends on the leading coefficient of the quadratic equation it is representing. The general features of a parabola include its axis of symmetry, its vertex, 𝑦-intercept, and π‘₯-intercepts. Let’s clear some space to review each of these features. To be a parabola, a curve must have perfect symmetry over a vertical line, which we call the axis of symmetry. The axis of symmetry for the graph that we’ve been given is the vertical line represented by the equation π‘₯ equals zero, or in other words the 𝑦-axis.

The next identifiable feature of a parabola is its vertex. This is the turning point found on the axis of symmetry. The vertex of a quadratic function will either be its minimum point or maximum point, depending whether the parabola curves upward or curves downward. According to this definition, we see that the vertex of the parabola we’ve been given is the point zero, four. The 𝑦-intercept of any function is the value of 𝑦 when π‘₯ equals zero. Considering our axis of symmetry is the equation π‘₯ equals zero, we will take the 𝑦-value of the vertex to be our 𝑦-intercept.

The final general feature of a parabola would be the π‘₯-intercepts, which are the values of π‘₯ when 𝑦 equals zero. There may be no π‘₯-intercepts, possibly one, or two π‘₯-intercepts. We recall that the equation 𝑦 equals zero is the π‘₯-axis and the equation π‘₯ equals zero is the 𝑦-axis. This means that we can think of the 𝑦-intercept as the point where the graph crosses the 𝑦-axis and the π‘₯-intercepts as the points where the graph crosses the π‘₯-axis. As shown by the two pink π‘₯’s, our graph crosses the π‘₯-axis at two distinct points: negative two, zero and two, zero. These two coordinate points are the π‘₯-intercepts of our quadratic function.

We were told that our equation is 𝑦 equals 𝑓 of π‘₯. And we are asked to find the solution set of the equation 𝑓 of π‘₯ equals zero. In other words, this means that we are looking for the solution set when 𝑦 equals zero. It should not be surprising to realize that the solution set relates back to the π‘₯-intercepts since these are the points at which the function has a 𝑦-value equal to zero.

We will now clear some space in order to review the definition of a solution set. We recall that the solution set 𝑆 of the function 𝑦 equals 𝑓 of π‘₯ is the set of all values of π‘₯ such that 𝑓 of π‘₯ equals zero. In other words, these are the π‘₯-values at which the graph crosses the π‘₯-axis. Those are the π‘₯-intercepts. These π‘₯-values are called the roots, the zeros, or the solutions of the function 𝑓 of π‘₯. There are three distinct possibilities for what could be contained in the solution set.

We will clear some space and review these three possibilities. The first possible outcome is when there are two real roots of the function, when π‘₯ equals 𝑆 sub one and π‘₯ equals 𝑆 sub two, where 𝑆 sub one is not equal to 𝑆 sub two. In this case, the solution set 𝑆 has two elements, 𝑆 sub one and 𝑆 sub two. The second possible outcome is that there is one real root of our function when π‘₯ equals 𝑆 sub one. In this case, the solution set has one element, 𝑆 sub one. Finally, it is possible that our function has no real roots. In this case, we say that 𝑆 is equal to the empty set.

When we have the graph of 𝑓 of π‘₯, it is actually quite straightforward to determine which of these possible outcomes is the case. A solution set with two elements corresponds to a parabola that crosses the π‘₯-axis twice. A solution set with one element corresponds to a parabola that crosses the π‘₯-axis at one point. In this unique case, the π‘₯-intercept is the same as the vertex of that parabola, making it either the minimum or maximum point of that graph. A solution set with no elements corresponds to a parabola that is either above or below the π‘₯-axis but never intercepts it at all.

We conclude that our function has two real roots because the parabola intercepts the π‘₯-axis at two points. As we determined earlier, 𝑓 of π‘₯ equals zero means 𝑦 equals zero, and this happens at two locations. The π‘₯-values from these two coordinate points are the two elements of our solution set. In conclusion, the solution set of the equation 𝑓 of π‘₯ equal to zero contains the two real roots of the function 𝑦 equals 𝑓 of π‘₯ as shown in the graph that we were given. And these two real roots were negative two and positive two. So those are the elements of our solution set.

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