Video Transcript
Which of the following graphs represents the equation π¦ is equal to π₯ plus four all squared?
In this question, weβre given five graphs, and we need to determine which of these five graphs represents the equation π¦ is equal to π₯ plus four all squared. And thereβs several different ways we could go about this. For example, we could try eliminating options. However, weβre instead going to try and sketch the curve π¦ is equal to π₯ plus four all squared. Since itβs useful to be able to sketch curves of this form and while not necessary to answer this question, we wonβt always be given the graphs of these functions. Now letβs recall how we sketch curves of this form.
We can see this is the graph of a quadratic equation, since this is a polynomial and the highest π₯ power is two. In fact, thereβs something even more useful we could notice. This equation is given in vertex form. Thatβs the form π¦ is equal to π times π₯ minus β all squared plus π, where π, β, and π are real numbers and π is not equal to zero. In this case, the coefficient of our parentheses is one. So, π is equal to one. Weβre adding four to our value of π₯, so our value of β is negative four. And we have no constant at the end; our value of π is zero. And the vertex form of a quadratic graph is very useful because it allows us to find the coordinates of its vertex. Thatβs the turning point.
And we recall the coordinates of a vertex written in the form π times π₯ minus β all squared plus π is the coordinates β, π. And since our value of β is negative four and π is zero, the coordinates of the vertex will be negative four, zero. To sketch the graph of this function, weβll add this coordinate to a pair of axes. And we recall this is not the only thing we can use the values of π, β, and π to determine about our graph. We can also determine whether our parabola will open upwards or downwards by using the sign of π. If π is positive, our parabola will open upwards and if π is negative, our parabola will open downwards. And in this case, our value of π is one, which is positive. So our parabola needs to open upwards. This gives us the following sketch.
And although this is enough to answer our question, we should also find the coordinates of one extra point on our curve. And this is because thereβs an infinite number of parabolas with vertex of coordinates negative four, zero, which open upwards. If we find the coordinates of one extra point on our curve, we have a unique parabola. Weβll find the coordinates of the π¦-intercept. Weβll do this by substituting π₯ is equal to zero into our curve. We get that π¦ is equal to zero plus four all squared, which simplifies to give us four squared, which is 16. And we can see that this matches the answer in option (B).
Therefore, we were able to show of the five given options only option (B) was a graph of the equation π¦ is equal to π₯ plus four all squared.