WEBVTT
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For the function π of π₯ equals π₯ squared minus four π₯ plus three, answer the following questions.
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Firstly, find by factoring the zeroes of the function.
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Secondly, identify the graph of π.
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There are also two further parts to this question.
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So firstly, weβre asked to find the zeros of this function.
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And the method weβre told to use is factoring.
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We therefore need to write our quadratic as the product of two linear factors.
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As the coefficient of π₯ squared is one, we know that the first term in each of our parentheses will be π₯.
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Weβre then looking for two numbers whose sum is the coefficient of π₯, thatβs negative four, and whose product is the constant term, thatβs positive three.
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Well, the two numbers that fit both of those criteria are negative one and negative three.
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Negative one plus negative three is negative four, and negative one multiplied by negative three is positive three.
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So our quadratic factors as π₯ minus one multiplied by π₯ minus three, which we can of course confirm by redistributing the parentheses if we wish.
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We need to use this factored form to determine the zeroes of the function, which we recall are the π₯-values such that π of π₯ equals zero.
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If we set this factored form equal to zero, we then recall that for the product of two things to be zero, at least one of them must themselves be zero.
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So we can take each factor in turn and set it equal to zero, giving two simple linear equations.
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The first can be solved by adding one to each side to give π₯ equals one, and the second can be solved by adding three to each side to give π₯ equals three.
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The roots or zeros of this function then are the values one and three.
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Now, in the second part of the question, weβre asked to identify the graph of our function π.
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And we can see that weβve been given three possibilities: a blue one, a red one, and a green one.
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Now, weβve just found that our graph has zeros at one and three.
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And remember, these zeros are the values of π₯ at which the graph crosses the π₯-axis.
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So if our graph crosses the π₯-axis at one and three, we can see from the figure that this leaves just the red and green graphs.
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The blue graph crosses the π₯-axis or has zeros at values of negative one and negative three.
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Now, we just need to decide between the red and green graphs, which we see are mirror images of each other.
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One is an upward-curving parabola, and the other is a downward-curving parabola.
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We recall that the type of parabola we have will be determined by the value of π.
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Thatβs the coefficient of π₯ squared.
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In our function, the coefficient of π₯ squared is one.
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Itβs a positive value, which means the parabola will curve upwards.
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That means then that the graph of our function π must be the red graph.
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It has the correct zeros and the correct shape.
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We can also see that the π¦-intercept of this graph is three, which is indeed the constant term in our function π of π₯.
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The remaining two parts of the question, which I didnβt write down initially because they give the game away for the previous part are.
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Write the equation for π, the function that describes the blue graph.
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And write the equation for β, the function that describes the green graph.
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Letβs look at this blue graph first of all then.
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We already said that it has zeros at negative one and negative three.
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This means that in its factored form, it has factors of π₯ plus one and π₯ plus three.
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But there could also be a factor of π that we multiply by.
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To determine whether the value of π is one or something else, we consider the π¦-intercept of the graph, which we can see is the same as the π¦-intercept of the red graph.
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Itβs three.
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When we multiply these two factors together, the constant term will be one multiplied by three, which is indeed three.
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And so this tells us that the value of π is simply one.
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Our function π in its factored form then is π₯ plus one multiplied by π₯ plus three.
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If we distribute the parentheses, we have π of π₯ equals π₯ squared plus four π₯ plus three.
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For the green graph, it has the same zeros as our function π.
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So it can be written as π multiplied by π₯ minus one multiplied by π₯ minus three.
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And again, we need to determine whether the value of π is one or something else.
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Well, the π¦-intercept for the green graph is negative three.
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If we multiply together negative one and negative three, we get a value of positive three.
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And so in order to ensure the π¦-intercept, the constant term in the expanded form of β of π₯, is negative three, we need the value of π to be negative one.
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The equation β of π₯ then is negative π₯ minus one multiplied by π₯ minus three.
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In fact, it is the complete negative of our function π of π₯, which we can also see because they are reflections of one another in the π₯-axis.
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We can write the equation β of π₯ then as the complete negative of our function π of π₯.
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β of π₯ is equal to negative π₯ squared minus four π₯ plus three.