WEBVTT
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Let us consider the function π of π₯ is equal to eight π₯ minus 11.
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Fill in the table.
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Identify the three points that lie on the line π¦ is equal to eight π₯ minus 11.
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There are two parts to this question.
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Firstly, we need to complete the table by filling in the values of π¦, or π of π₯, for the function eight π₯ minus 11.
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In the top row of our table, weβre given π₯-values of negative one, zero, and one.
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And we can calculate the π¦-values in the second row of our table by substituting the corresponding π₯-values into our function.
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Letβs begin with negative one.
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π of negative one is equal to eight multiplied by negative one minus 11.
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And this is equal to negative 19.
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The first missing number in our table is negative 19.
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Repeating this for π₯ equals zero, we have π of zero is equal to eight multiplied by zero minus 11, which is equal to negative 11.
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When π₯ is equal to zero, π of π₯, or π¦, is equal to negative 11.
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Finally, we need to calculate the π¦-value when π₯ equals one.
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Eight multiplied by one minus 11 is equal to negative three.
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The three missing values in the table are negative 19, negative 11, and negative three.
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The second part of our question asks us to identify the three points on the graph that lie on the line π¦ is equal to eight π₯ minus 11.
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The easiest way to do this is to consider the three pairs of values we found in the first part of this question.
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These are the ordered pairs, or coordinates, negative one, negative 19; zero, negative 11; and one, negative three.
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And we can find the points on the graph by going along the horizontal or π₯-axis and then up or down the vertical π¦-axis.
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Going along to negative one and then down to negative 19, we find the point πΌ.
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In the same way, the point π» has coordinates zero, negative 11.
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And finally, πΊ has coordinates one, negative three.
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The three points that lie on the line π¦ is equal to eight π₯ minus 11 are πΌ, π», and πΊ.
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Since our equation is written in the form π¦ is equal to ππ₯ plus π, we know it is a linear equation.
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We will have a straight line with slope, or gradient, equal to negative eight and a π¦-intercept of negative 11.
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Drawing this straight line, we see that it does indeed pass through the points πΌ, π», and πΊ.