WEBVTT
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Which of the following graphs represents the equation π¦ is equal to π₯ squared plus three?
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So in order to enable us to solve this problem, what Iβve done is that Iβve actually drawn a sketch of the graph π¦ is equal to π₯ squared.
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As we can see with π¦ is equal to π₯ squared, we actually have a repeating root.
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So it just touches the π₯-axis.
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And it does that at the origin, so the point zero, zero.
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And thatβs because if π¦ is equal to zero, then π₯ will also be equal to zero.
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We can also note that the π¦ equals π₯ squared graph is a U-shaped parabola.
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Okay, brilliant.
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So we know the basic shape of π¦ equals π₯ squared.
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Letβs look at ours which is π¦ is equal to π₯ squared plus three.
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So our graph is going to be a U-shaped parabola, like π¦ is equal to π₯ squared.
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But because itβs π¦ is equal to π₯ squared plus three, we know that transformation is taking place.
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And this transformation is a shift.
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And weβll have a look at some shift rules to work out which shift it is.
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Our first shift rule is that π π₯ plus π, and the plus π is within the parenthesis, is equal to a shift of negative π units in the π₯-axis.
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What this actually means is that all the π₯-coordinates will actually have π subtracted from them.
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And the other shift rule is that π π₯ plus π, this time noting that the plus π is actually outside the parenthesis, is equal to a shift of π units in the π¦-axis.
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And what this means in practice is actually we add π units to all of our π¦-coordinates.
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So if we take that back at our function, weβve got π¦ is equal to π₯ squared plus three.
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Well, this is actually just like the second shift rule that weβve looked at because itβs the same as that π₯ squared.
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And then weβve added three to it.
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So therefore, our transformation is going to be a shift of plus three units in the π¦-axis.
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So what this means in practice is that weβre actually going to add three to the π¦-coordinates of the minimum point cause this is gonna help us identify the graph.
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Well, our minimum point in a graph of π¦ equals π₯ squared was zero, zero.
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So therefore, our minimum point is gonna be equal to zero, three because weβve added three units onto our π¦-coordinate.
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Okay, great.
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So with this in mind, letβs choose which graph is the correct graph of this equation.
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So therefore, we can say that the graph π is the correct graph that represents the equation π¦ is equal to π₯ squared plus three.
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And we can see that because the minimum point, as we highlighted here, is zero, three.
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And thatβs what weβre looking for because weβve actually shifted the π₯ squared graph three units in the π¦-axis.