Question Video: Finding the Values of Unknowns of an Absolute Value Function from Its Graph | Nagwa Question Video: Finding the Values of Unknowns of an Absolute Value Function from Its Graph | Nagwa

# Question Video: Finding the Values of Unknowns of an Absolute Value Function from Its Graph Mathematics

The graph in figure (i) is of π(π₯) = (5/2) |π₯| + (1/2 π₯), which could also be written as follows π(π₯) = 3π₯, π₯ β₯ 0 and π(π₯) = β2π₯, π₯ < 0. Find the values of π and π that would make graph (ii) that of π(π₯) = π|π₯ β 5| + π(π₯ β 5).

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### Video Transcript

The graph in figure (i) is of π of π₯ is equal to five over two times the absolute value of π₯ plus one-half π₯, which could also be written as follows. π of π₯ is equal to three π₯ when π₯ is greater than or equal to zero and π of π₯ is equal to negative two π₯ when π₯ is less than zero. Find the values of π and π that would make graph (ii) that of π of π₯ is equal to π times the absolute value of π₯ minus five plus π times π₯ minus five.

In this question, weβre given the graphs of two functions, and weβre given two different ways of expressing the function in the first graph. We can either represent it by using an absolute value symbol or we can represent it by using a piecewise-defined function. We can see that the second graph is very similar to the first graph. In fact, weβre told that π of π₯ can be written in a similar manner to π of π₯, by using an absolute value symbol. We need to use all of this information to determine the values of π and π, and there are many different ways of going about doing this. For example, we can do this directly from the diagram and the definition of π of π₯. However, weβre first going to look up both definitions of π of π₯ and see how this relates to the diagram.

Letβs start by looking at the piecewise definition of π of π₯. We can look at the first subfunction of π of π₯. Thatβs three π₯ when π₯ is greater than or equal to zero. Of course, this is the straight line of slope three passing through the origin where we only plot when π₯ is greater than or equal to zero. We can mark this on the diagram in orange. We can then do the same for the second subfunction. Itβs marked in pink. Therefore, we could determine the piecewise definition of our function π of π₯ by determining the equations of each of the subfunctions. From the diagram, we could determine that the pink line has equation negative two π₯ and the orange line has equation three π₯. This would then give us the piecewise definition of π of π₯.

We could do this now for our function π of π₯ by using this diagram. However, this would not directly give us the values of π and π. Instead, we need to determine how the values of π and π will be related to the piecewise definition. To do this, letβs once again look at our function π of π₯. However, this time weβll look at the absolute value definition of π of π₯. Since the absolute value of π₯ changes depending on whether π₯ is positive or negative, we can see what happens in both of these cases. First, if π₯ is greater than or equal to zero, then the absolute value of π₯ is just equal to π₯. Therefore, π of π₯ will be equal to five π₯ plus a half π₯, which we can evaluate as three π₯, which is of course the first subfunction of π of π₯.

We can do exactly the same when π₯ is less than zero. In this case, the absolute value of π₯ will be negative π₯, so π of π₯ is negative five over two π₯ plus one-half π₯, which is negative two π₯, the second subfunction of π of π₯. Therefore, in the case of π of π₯, the sum of the two coefficients that would be the values of π and π for π of π₯ gave us the slope of the first subfunction, and then flipping the sign of the first coefficient and adding the second coefficient gave us the slope of the second subfunction. This is a relationship between the equations of each subfunction and the coefficient values weβre looking for.

We can now apply this same process to π of π₯. And there are many different ways we could go about this. For example, we could find a piecewise definition of π of π₯ from its diagram. However, itβs not necessary. Instead, weβre just going to do this directly from the equation weβre given. First, letβs start when π₯ is greater than or equal to five. In this case, the absolute value of π₯ minus five will just be π₯ minus five because itβs positive. This gives us that π of π₯ is equal to π times π₯ minus five plus π times π₯ minus five when π₯ is greater than or equal to five. And weβve marked the section of our graph when π₯ is greater than or equal to five in blue.

We can then distribute over the parentheses and simplify this expression. We get π plus π multiplied by π₯ minus five π minus five π. This will be the subfunction for the section marked in blue on the diagram. And now thereβs a few different ways we can use this to find an expression in terms of π and π. Weβre going to use the fact that this is a linear function. And in a linear function, the coefficient of π₯ is the slope of the line. This means the slope of the blue line needs to be equal to π plus π, and we can find the slope of the line from the diagram. For every five units we go across, we move five units up. The slope of the line is one. Therefore, the coefficient of π₯ needs to be one. π plus π is equal to one.

Letβs now follow the same process for the other subfunction. In this subfunction, our values of π₯ are less than five, so the absolute value of π₯ minus five will be equal to five minus π₯. π of π₯ is π times five minus π₯ plus π times π₯ minus five. We can then distribute and simplify in exactly the same way. This time, we get π minus π multiplied by π₯ plus five π minus five π. Now we can find an expression for π minus π by equating the slopes of the lines. We can then determine the slope of this line from the diagram. For every five units we move right, the line moves 15 units down. And negative 15 divided by five is negative three. The slope of the line is negative three. So the coefficient of π₯ is negative three. π minus π is equal to negative three.

Therefore, we have two equations in π and π. We can solve these simultaneously, and thereβs several different ways of doing this. Weβre going to add the two equations together to eliminate π. We get that two π will be equal to negative two, which we can solve for π by dividing through by two. π is negative one. Finally, we can determine the value of π by substituting our value of π into one of our previous equations. For example, we know π plus π is equal to one and π is negative one, so our value of π is two.

Therefore, we were able to show for the graph in figure (ii) to be the graph of π of π₯ is equal to π times the absolute value of π₯ minus five plus π times π₯ minus five, the value of π must be two and the value of π must be negative one.