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In this video, we will learn how to evaluate and graph absolute value functions and identify their domain and range.
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We will begin by recalling what we mean by an absolute value function.
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An absolute value function contains an algebraic function within absolute value symbols.
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The absolute value of any number is its distance from zero on the number line.
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Letβs consider the function π of π₯ which is equal to the absolute value, sometimes known as the modulus, of π₯.
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To graph this function, we will choose some integer values of π₯ and find some ordered pairs.
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The absolute value of negative two is equal to two as it is two away from zero on the number line.
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The absolute value of negative one is equal to one as this is one away from zero on the number line.
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This leads us to the fact that the absolute value of any number cannot be negative.
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We are not interested in the sign and just the distance of the number from zero.
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The absolute value of zero, one, and two are zero, one, and two, respectively.
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We can then graph this function on the coordinate plane where π¦ is equal to π of π₯.
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Our first coordinate or ordered pair is negative two, two.
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We then have negative one, one.
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Our other three points are zero, zero; one, one; and two, two.
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Connecting these points, we create a V-shaped graph.
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This will be true for the absolute value of any linear function of the form ππ₯ plus π.
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Letβs now consider some key points or information from this graph.
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The vertex of the graph has coordinates zero, zero.
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This is the minimum point of the function the absolute value of π₯.
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The π¦-axis is a line of symmetry.
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This means that the equation of the line of symmetry is π₯ equals zero.
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We know that the domain of any function is those values that can be inputted into the function.
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As we can substitute any value of π₯ into our function, the domain is the set of all real numbers.
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This can also be written as the open interval negative β to the open interval β.
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The range is the outputs of the function.
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It is all the values of π¦ or π of π₯.
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These will all lie above or on the π₯-axis.
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Therefore, π of π₯ is greater than or equal to zero.
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Once again, this can be written as an interval, from the closed interval zero to the open internal β.
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Finally, we see that the π¦-intercept and π₯-intercept are both zero.
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We will now look at a couple of questions where we need to find the domain and range of an absolute value function from its graph.
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Find the range of the function π of π₯ is equal to the absolute value of negative two π₯ minus two.
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The range of a function π¦ equals π of π₯ is the set of values π¦ takes for all values of π₯ within the domain of π.
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For any graph drawn on the coordinate plane, the range or output is all the set of values that π¦ takes.
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For the function π of π₯ which is equal to the absolute value of negative two π₯ minus two, all our values of π¦ are above or on the π₯-axis.
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We can therefore say that π¦ or π of π₯ must be greater than or equal to zero.
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This can be written using interval notation from the left-closed, right-open interval zero to β.
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We have a closed interval at zero as the function can equal zero.
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This is denoted by the closed circle or dot at negative one, zero.
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As π of π₯ can never actually reach β, we use an open interval for the upper limit.
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Determine the domain and range of the function π of π₯ is equal to the absolute value of negative π₯ minus one plus one.
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The domain of any function is the set of all possible input values.
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When the graph is drawn on the coordinate plane, this is denoted by all the π₯-values that can be substituted into the function.
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As we can substitute any value of π₯ into our function, the domain is the set of all real values.
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This can also be written as the set of values between the open interval negative β and the open interval β.
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The range of a function is the set of all possible output values.
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On the coordinate plane, it is all the values of π¦ or π of π₯.
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We can see on our graph that the vertex or minimum point is at negative one, one.
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The range of the function π is therefore from the left-closed, right-open interval one to β.
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The solid circle or dot at negative one, one means that we will use the closed interval as the value of one is included in our range.
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The function π of π₯ which is the absolute value of negative π₯ minus one plus one has a domain of all real values and a range from one to β.
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In our next two questions, we will calculate the domain and range without being given a graph.
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Given that π is a constant, what is the domain of the function π of π₯ is equal to the absolute value of π₯ plus π?
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Letβs begin by recalling what the graph of the function the absolute value of π₯ looks like.
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This is a V-shaped graph with minimum points or vertex at zero, zero.
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We recall that the domain is the set of input or π₯-values.
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This means that the domain of π of π₯, the absolute value of π₯, is all real values.
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Whilst it is not mentioned in this question, the range is the set of π¦ or output values.
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As the π¦-values of π of π₯ are all greater than or equal to zero, the range of π of π₯ is from the left-closed, right-open interval from zero to β.
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Letβs now consider our function π of π₯, which is equal to the absolute value of π₯ plus π.
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The function π of π₯ is a horizontal translation of π of π₯ π units to the left.
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If π is a positive number, the graph will shift or translate π units to the left.
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If π on the other hand was a negative number, the graph would shift or translate to the right.
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As the graph has just shifted horizontally, the range and domain have not altered.
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The domain of the function π of π₯ which is equal to the absolute value of π₯ plus π is the set of all real values.
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This could also be written as the set of values from negative β to β.
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Find the domain and range of the function π of π₯ is equal to negative four multiplied by the absolute value of π₯ minus five minus one.
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Letβs begin by looking at what the general function the absolute value of π₯ looks like.
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This is a V-shaped graph with minimum point or vertex at zero, zero.
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We will now consider the transformations that are made to obtain the function π of π₯.
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Letβs begin by considering the absolute value of π₯ minus five.
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This is a translation five units to the right.
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This function would therefore have a minimum point or vertex at five, zero.
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Multiplying the absolute value of π₯ minus five by negative four results in a stretch of scale factor negative four.
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This means that the graph will be four times steeper and also reflected in the π₯-axis.
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Finally, we need to subtract one from this function.
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This will result in a translation one unit down in the π¦-direction.
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The function π of π₯ which is equal to negative four multiplied by the absolute value of π₯ minus five minus one is shown in green.
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We will now remove some of the other graphs from the coordinate plane.
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The vertex or maximum point of π of π₯ is at the coordinates five, negative one.
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We know that the domain of any function is the set of π₯- or input values.
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As we can input any value into our function π of π₯, the domain is the set of all real values from negative β to β.
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The range is the set of all output or π¦-values.
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From the graph, we can see that these are all values less than or equal to negative one.
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The range of π of π₯ is therefore equal to the set of values on the left-open, right-closed interval from negative β to negative one.
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In the final question in this video, we will evaluate an absolute value function by direct substitution.
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A body was moving with a uniform velocity of magnitude five centimeters per second from the point π΄ to the point πΆ passing through the point π΅ without stopping.
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The distance between the body and point π΅ is given by π of π‘ is equal to five multiplied by the absolute value of eight minus π‘, where π‘ is the time in seconds and π is the distance in centimeters.
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Determine the distance between the body and the point π΅ after five seconds and after 11 seconds.
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We are given a diagram which shows the body that is about to move from point π΄ to point πΆ via point π΅ with a velocity of five centimeters per second.
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Whilst there is a lot of information in this question, the key point is that the function π of π‘ is equal to five multiplied by eight minus π‘.
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π of π‘ is the distance of the body from point π΅ after a given time.
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We need to calculate this distance after five seconds and also after 11 seconds.
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After five seconds, π‘ is equal to five.
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Therefore, we need to calculate π of five.
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This is equal to five multiplied by the absolute value of eight minus five.
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Eight minus five is equal to three, so we need to multiply five by the absolute value of three.
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As the absolute value of a number is its distance from zero, the absolute value of three is three.
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As five multiplied by three is equal to 15, the distance between the body and the point π΅ after five seconds is 15 centimeters.
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We need to repeat this process when π‘ equals 11.
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This means we need to calculate the value of π of 11.
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This is equal to five multiplied by the absolute value of eight minus 11.
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Eight minus 11 is equal to negative three.
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As the absolute value of a number is its distance from zero, the absolute value of negative three is also three.
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In fact, the absolute value of any number will always be positive.
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Multiplying five by three once again, we see that the distance between the body and point π΅ after 11 seconds is also 15 centimeters.
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In terms of our diagram, we can see that after five seconds and 11 seconds, the body is the same distance away from point π΅.
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After five seconds, it is still approaching point π΅ from point π΄.
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And after 11 seconds, it is past point π΅ and itβs heading towards Point πΆ.
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We will now summarize the key points from this video.
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In this video, we saw that the absolute value of a number is its distance from zero.
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This means that the absolute value of a number can never be negative.
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The domain of any function is the set of π₯- or input values.
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We saw that when dealing with the absolute value of linear functions of the form ππ₯ plus π, the domain was equal to all real values from the open interval negative β to the open interval β.
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The range of a function is the set of π¦- or output values.
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If π of π₯ is equal to the absolute value of ππ₯ plus π, then the range will be from the closed interval zero to the open interval β.
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The range will contain all values of π¦ greater than or equal to zero.
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We also saw in this video that the transformations of the absolute value of linear functions sometimes alter the range but never alter the domain.
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We saw that the function π of π₯ which is equal to π of π₯ minus β translates the function π of π₯ equals the absolute value of π₯ horizontally.
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It shifts the graph β units to the right.
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π of π₯ plus β would move the graph β units to the left.
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Likewise, the function π of π₯ which is equal to π of π₯ plus π translates the function π of π₯ equals the absolute value of π₯ vertically.
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This translation shifts the graph π units up.
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As with the previous example, π of π₯ minus π this time would shift the graph π units down.
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Finally, we saw that the function π of π₯ which is equal to π multiplied by π of π₯ stretches the function π of π₯ equals the absolute value of π₯.
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The value of π is the scale factor.
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And if a is π negative number, the graph is reflected in the π₯-axis.
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This means that it opens downwards instead of upwards.
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The V-shaped graph is now upside down.
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The last two transformations will have an impact on the range of the absolute value function.