The Cubic Curve 1
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Question 1 of 5
1. Question
Write the equation of the cubic graph that is shown on the number plane below.
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This cubic curve does not have any turning points and it passes through the origin. Therefore it is in the form `y=ax^3`.Use the point `(1,4)` shown on the graph to find the value of `a`.The `x`-coordinate of this point is `1`, and the `y`-coordinate of this point is `4`.So we should substitute `x=1` and `y=4` into the equation `y``=a``x^3`.`4``=a(``1^3``)`.Solve this equation to find the value of `a`.`a=4`Substitute the value for `a` and write the equation of the graph.`y=4x^3` -
Question 2 of 5
2. Question
Sketch the graph of the cubic curve for the equation below`y=2(x+1)(x-4)(x-6)`Hint
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The standard form for graphing a cubic is `y=a(x-r)(x-s)(x-t)`.The sign of `a` in this equation tells us the shape of the graphIn this equation `a=2` which is positive.Therefore this is a positive cubic curve and has the approximate shape shown below:
Find the `x`-intercepts of the graph by setting `y=0`.`2``(x+1)``(x-4)``(x-6)``=0`So `x+1=0`, `x-4=0` or `x-6=0`This means `x=-1`, `x=4` or `x=6`.This gives us `x`-intercepts `-1`, `4` and `6` which we mark on the `x`-axis
Find the `y`-intercepts of the graph by setting `x=0`.`y=2(0+1)(0-4)(0-6)``y=2\cdot (1) \cdot (-4)\cdot (-6)``y=48`This gives us a `y`-intercept of `48` which we mark on the `y`-axis
Draw a positive cubic curve which passes through the intercepts found.
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Question 3 of 5
3. Question
Sketch the graph of the cubic curve for the equation below`y=-2(x-1)(x+2)(x+4)`Hint
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The standard form for graphing a cubic is `y=a(x-r)(x-s)(x-t)`.The sign of `a` in this equation tells us the shape of the graphIn this equation `a=-2` which is negative.Therefore this is a negative cubic curve and has the approximate shape shown below:
Find the `x`-intercepts of the graph by setting `y=0`.`-2``(x-1)``(x+2)``(x+4)``=0`So `x-1=0`, `x+2=0` or `x+4=0`This means `x=1`, `x=-2` or `x=-4`.This gives us `x`-intercepts `1`, `-2` and `-4` which we mark on the `x`-axis
Find the `y`-intercepts of the graph by setting `x=0`.`y=-2(0-1)(0+2)(0+4)``y=-2\cdot (-1) \cdot (2)\cdot (4)``y=16`This gives us a `y`-intercept of `16` which we mark on the `y`-axis
Draw a negative cubic curve which passes through the intercepts found.
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Question 4 of 5
4. Question
Write the equation of the cubic graph that is shown on the number plane below.
Hint
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This cubic curve does not have any turning points and it passes through the origin. Therefore it is in the form `y=ax^3`.Use the point `(-2,16)` shown on the graph to find the value of `a`.The `x`-coordinate of this point is `-2`, and the `y`-coordinate of this point is `16`.So we should substitute `x=-2` and `y=16` into the equation `y``=a``x^3`.`16``=a``(-2)^3`.Solve this equation to find the value of `a`.`16=-8a``\frac{16}{-8}=a``a=-2`Substitute the value for `a` and write the equation of the graph.`y=-2x^3` -
Question 5 of 5
5. Question
Write the equation of the cubic graph that is shown on the number plane below.
Hint
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This cubic curve does not have any turning points and it passes through the origin. Therefore it is in the form `y=ax^3`.The value of `d` will be given by the `y`-intercept.The graph crosses the `y`-axis at `(0,``5``)`, therefore `d=5`.The equation of the graph has the form `y=ax^3 ``+5`.Use the point `(2,-3)` shown on the graph to find the value of `a`.The `x`-coordinate of this point is `2`, and the `y`-coordinate of this point is `-3`.So we should substitute `x=2` and `y=-3` into the equation `y``=a``x^3``+5`.`-3``=a``(2^3)``+5`.Solve this equation to find the value of `a`.`-3=8a+5``-8=8a``\frac{-8}{8}=a``a=-1`Substitute the value for `a` and write the equation of the graph.`y=-x^3+5`