### Video Transcript

Consider the equation π¦ is equal to π₯ squared minus five π₯ plus three. In the following, find a solution by filling in the blank space. Part (1) the coordinates two and what, part (2) the coordinates negative one and what, part (3) the coordinates zero and what, and part (4) the coordinates four and what. These solutions π₯, π¦ can be plotted as seen in the figure. What will be true of any other solution plotted in the plane? Option (A) they will lie in the same quadrant. Option (B) they will not lie in the same quadrant. Option (C) they will not lie on the curve. Option (D) they will lie on the π₯-axis. Or option (E) they will lie on the curve.

In this question, weβre considering a given quadratic equation: π¦ is equal to π₯ squared minus five π₯ plus three. And weβre even given a sketch of this graph. We need to use both of these informations to answer five questions about this equation. The first four parts are about finding solutions to this equation. Thatβs values of π₯ and π¦ which satisfy the equation. And we can start by recalling in an ordered pair such as this, the first value represents the π₯-value and the second value will represent the π¦-value.

So the first part of this question wants us to determine the value of π¦ when π₯ is equal to two. We can substitute π₯ is equal to two into the quadratic equation. We get π¦ is equal to two squared minus five times two plus three. We can then evaluate this expression. Two squared is four, and five times two is 10. So we get four minus 10 plus three, which is equal to negative three. Therefore, weβve shown if π₯ is equal to two, π¦ must be equal to negative three. This is the missing value in our ordered pair for the solution. Therefore, the answer to the first part of this question is two, negative three.

We can follow the same process to find the other three missing values. Letβs start with the second part. Our value of π₯ is negative one. We substitute this into our quadratic equation to get π¦ is equal to negative one squared minus five times negative one plus three, which we can then evaluate. Negative one squared is one, and negative five times negative one is equal to five. So we get one plus five plus three, which we can calculate is nine. Therefore, weβve shown when π₯ is negative one, our value of π¦ needs to be equal to nine. So the answer to the second part of this question is negative one, nine.

We can continue this process for the third part of our question. We substitute π₯ is equal to zero into the quadratic equation. We get that π¦ will be equal to zero squared minus five times zero plus three. And both of the first two terms have a factor of zero. So this evaluates to give us three. Therefore, the solution to the third part of this question is zero, three. And finally, for the fourth part of this question, we substitute π₯ is equal to four into the quadratic equation. We get π¦ is equal to four squared minus five times four plus three, which evaluates to give us negative one, giving us the answer to the fourth part of this question is four, negative one.

In the next part of this question, we plot the ordered pairs π₯ and π¦ on a pair of axes. And what this means is we plot every solution to the equation π¦ is equal to π₯ squared minus five π₯ plus three. Therefore, the π₯- and π¦-coordinates of any point which lie on this curve must be a solution to the quadratic equation. For example, because the point with coordinates five, three lies on the curve, we can conclude π₯ is equal to five and π¦ is equal to three will solve the quadratic equation. And this is a direct result of plotting all of the points which solve the equation. And this is enough to answer our question. We need to determine what will be true of any other solution plotted in the plane.

Since the curve contains all of the solutions to the quadratic equation, we can conclude that any other solution must also lie on the curve. And this is given as option (E) they will lie on the curve.