Question Video: Matching the Rule of a Quadratic Function with Its Graph Mathematics • Third Year of Preparatory School

Video Transcript

Which of the following graphs represents 𝑓 of 𝑥 equals negative one times 𝑥 minus one squared plus two? Graphs (a), (b), (c), and (d).

In this question, we are given a quadratic function 𝑓 of 𝑥 and asked to determine which of four given graphs represents the graph of this function. There are many different ways we can answer this question; we will go through two of these. Let’s start by determining information about the graph of 𝑓 of 𝑥 from its definition.

We can start by noting that 𝑓 of 𝑥 is a quadratic function, so we know the shape of its graph will be a parabola. We can also note that if we expanded the expression, we would find that the coefficient of 𝑥 squared is negative, so the parabola must open downwards. This allows us to eliminate graphs (a) and (d) since they open upwards.

We can determine which of the graphs (b) or (c) is correct by finding the coordinates of a point on the graph. We will check when 𝑥 equals zero, since we can see that both of these graphs have different 𝑦-intercepts. We substitute 𝑥 equals zero into the function 𝑓 of 𝑥 to obtain 𝑓 of zero equals negative one times zero minus one squared plus two. If we evaluate this expression, then we get one. This means that the 𝑦-intercept of the graph is at one. So, we can say that graph (c) is not possible since its 𝑦-intercept is at negative three. Hence, the correct graph must be graph (b).

This method requires us to be given options to eliminate in order to identify the correct graph. We can also sketch this graph by constructing a table of values. We have already calculated that 𝑓 evaluated at zero is equal to one. If we evaluate 𝑓 at one by substituting 𝑥 equals one into the function, we obtain two. Similarly, when 𝑥 equals two, we can calculate that 𝑓 of 𝑥 is equal to one.

We can follow this same process to fill in the rest of the values in this table. This gives us the coordinates of five points on the graph. We can then notice something interesting about the table of values we have constructed: the outputs of the function when 𝑥 equals zero and 𝑥 equals two are the same.

We can then recall that all parabolas have a line of symmetry through a point called their vertex. We can note that the line of symmetry will be halfway between these two points with the same output. This is when 𝑥 equals one. We can then use this to sketch the parabola passing through these points with vertex at the point one, two. We can then see that this matches graph (b).