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Write an Equation for Cubic CurvesWrite an Equation for Cubic Curves
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Question 1 of 7
1. 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 `(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 7
2. 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 3 of 7
3. 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,``4``)`, therefore `d=4`.The equation of the graph has the form `y=ax^3 ``4`.Use the point `(2,12)` 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 `12`.So we should substitute `x=2` and `y=12` into the equation `y``=a``x^3``4`.`12``=a``(2^3)``4`.Solve this equation to find the value of `a`.`12=8a4``16=8a``\frac{16}{8}=a``a=2`Substitute the value for `a` and write the equation of the graph.`y=2x^34` 
Question 4 of 7
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`.The value of `d` will be given by the `y`intercept.The graph crosses the `y`axis at `(0,``2``)`, therefore `d=2`.The equation of the graph has the form `y=ax^3 ``+2`.Use the point `(3,7)` shown on the graph to find the value of `a`.The `x`coordinate of this point is `3`, and the `y`coordinate of this point is `7`.So we should substitute `x=3` and `y=7` into the equation `y``=a``x^3``+2`.`7``=a``(3^3)``+2`.Solve this equation to find the value of `a`.`7=27a+2``9=27a``\frac{27}{9}=a``a=\frac{1}{3}`Substitute the value for `a` and write the equation of the graph.`y=\frac{1}{3}x^3+2` 
Question 5 of 7
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` 
Question 6 of 7
6. Question
The following equation is graphed below:
`y=a(x1)(x+1)(x3)`What is the value of `a`?Hint
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The standard form for graphing a cubic is `y=a(xr)(xs)(xt)`.Since the point `(``2``,``9``)` lies on the graph we can substitute `x=2` and `y=9` into the equation and solve to find `a`.`9``=a(``2``1)(``2``+1)(``2``3)``9=a(1)(3)(1)``9=3a``\frac{9}{3}=a``a=3` 
Question 7 of 7
7. Question
Write the equation of the cubic graph shown below.Hint
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The standard form for graphing a cubic is `y=a(xr)(xs)(xt)`, where `r`, `s`, and `t` are the `x`intercepts of the graph.Find the values of the `x`intercepts of the graph.One `x`intercept is at `1`, so `r=1`.The graph touches the `x`axis at `2`, meaning it will be a repeated root, so `s=2` and `t=2`Substitute the values for `r`, `s` and `t` into the standard equation`y=a(x``1``)(x(``2``))(x(``2``))``y=a(x1)(x+2)^2`Since the graph passes through `(``0``,``12``)`, substitute `x=0` and `y=12` into the equation and solve to find `a`.`12``=a(``0``1)(``0``+2)^2``12=a(1)(2)^2``12=4a``\frac{12}{4}=a``a=3`Substitute the value for `a` and write the complete equation.`y=3(x1)(x+2)^2`
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