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secondary 3 | A Maths
2 Answers Below
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help would be greatly appreciated asap thank you !!
Date Posted: 2 years ago
Views: 395
The expression in part ii is simply the reciprocal of the expression in part i.
Before I proceed to the main portion of part ii, I need to brief you on what's happening in the graph.
At the two points where the graph of y = 3x² + 12x - 5 meets the x-axis, y = 0; this will result in two corresponding values of x for which 3x² + 12x - 5 = 0.
Now, for the same values of x (i.e. the ones which result in 3x² + 12x - 5 becoming zero), what happens to the corresponding values of y if y = 1 / (3x² + 12x - 5)? Our values of y would become 1/0 which is undefined!
In other words, the graph of y = 1 / (3x² + 12x - 5) will not be valid for the two values of x.
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Now that we know this, we will zoom into just the portion in between these values of x (i.e. the portion of the original graph which is below the x-axis).
What happens when you divide 1 by a negative number? It becomes negative.
What happens when you divide 1 by a small negative number? It will become a large negative number. For example, 1/(-0.02) = -50.
What happens when you divide 1 by a large negative number? It will become a small negative number. For example, 1/(-50) = -0.02.
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You can see that for the original equation in part i, its lowest output value is -17.
This is the largest negative output amongst the negative portions of the graph.
When you take 1 divided by the largest negative output of -17, it will in fact become the smallest negative output of -1/17.
And for this reason, you will find that suddenly, the minimum point of the curve in part i becomes the maximum point of the curve in part ii.
I will attach a Desmos graph for you to understand what's happening more clearly.
Note that the positive portions of the graph is of little relevance to this question.
Before I proceed to the main portion of part ii, I need to brief you on what's happening in the graph.
At the two points where the graph of y = 3x² + 12x - 5 meets the x-axis, y = 0; this will result in two corresponding values of x for which 3x² + 12x - 5 = 0.
Now, for the same values of x (i.e. the ones which result in 3x² + 12x - 5 becoming zero), what happens to the corresponding values of y if y = 1 / (3x² + 12x - 5)? Our values of y would become 1/0 which is undefined!
In other words, the graph of y = 1 / (3x² + 12x - 5) will not be valid for the two values of x.
---------------------------------------------------------
Now that we know this, we will zoom into just the portion in between these values of x (i.e. the portion of the original graph which is below the x-axis).
What happens when you divide 1 by a negative number? It becomes negative.
What happens when you divide 1 by a small negative number? It will become a large negative number. For example, 1/(-0.02) = -50.
What happens when you divide 1 by a large negative number? It will become a small negative number. For example, 1/(-50) = -0.02.
---------------------------------------------------------
You can see that for the original equation in part i, its lowest output value is -17.
This is the largest negative output amongst the negative portions of the graph.
When you take 1 divided by the largest negative output of -17, it will in fact become the smallest negative output of -1/17.
And for this reason, you will find that suddenly, the minimum point of the curve in part i becomes the maximum point of the curve in part ii.
I will attach a Desmos graph for you to understand what's happening more clearly.
Note that the positive portions of the graph is of little relevance to this question.
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done
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Part i
I will type out part ii in the main question’s chat box.
I will type out part ii in the main question’s chat box.
Date Posted:
2 years ago
thank you very much helped a lot !!
done
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Here's the relevant curve I was mentioning; we only focus on the portion of the graph in between the two mentioned x values (I have included them here as blue and green "forbidden" lines).
Each time a number becomes more negative, its reciprocal becomes less negative. You can play around with actual numbers to understand this better.
This is the reason why our minimum point, which carried the most negative number possible, suddenly became the maximum point which carried the least negative number possible.
Each time a number becomes more negative, its reciprocal becomes less negative. You can play around with actual numbers to understand this better.
This is the reason why our minimum point, which carried the most negative number possible, suddenly became the maximum point which carried the least negative number possible.
Date Posted:
2 years ago
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