The Cubic Curve 2

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Question 1 of 4

1. Question
Write the equation of the cubic graph that is shown on the number plane below.
- `y=2x^3-4`
- `y=4x^3-2`
- `y=\frac{3}{2}x^3`
- `y=x^3-4`
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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=8a-4`

`16=8a`

`\frac{16}{8}=a`

`a=2`

Substitute the value for `a` and write the equation of the graph.

`y=2x^3-4`
Question 2 of 4

2. Question
Write the equation of the cubic graph that is shown on the number plane below.
- `y=-\frac{1}{3}x^3+2`
- `y=\frac{1}{3}x^3-2`
- `y=3x^3+2`
- `y=x^3+2`
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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 3 of 4

3. Question
The following equation is graphed below:
`y=a(x-1)(x+1)(x-3)`

What is the value of `a`?
- `3`
- `-3`
- `-\frac{9}{2}`
- `\frac{3}{2}`
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The standard form for graphing a cubic is `y=a(x-r)(x-s)(x-t)`.

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 4 of 4

4. Question
Write the equation of the cubic graph shown below.
- `y=-3(x-1)(x+2)^2`
- `y=3(x-1)(x-2)^2`
- `y=-3(x-1)(x+2)(x-2)`
- `y=3(x-1)^2(x+2)`
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The standard form for graphing a cubic is `y=a(x-r)(x-s)(x-t)`, 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(x-1)(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(x-1)(x+2)^2`