Q.Respected Sir/Madam;

Sketch the graph of `y=(x+2)^3 - 5`

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`y=(x+2)^3-5`

To graph this, transformation of function can be applied.

Since the given is a cubic function, let's start with the graph of the basic cubic function which is `y_0=x^3` .

Then, consider the expression inside the parenthesis `y_1=(x+2)^3` .

Notice that the variable x is added by 2. That means to graph `y_1=(x+2)^3` , the curve above `(y_0)` should be moved sidewards, which is 2 units to the left.

And, consider the number after `(x+2)^3` . We would then have `y=(x+2)^3 -5` .

Since `y_1` is equal to `(x+2)^3` , then the given function can be express as

`y=y_1-5` .

Here, notice that the variable `y_1` is subtracted by 5. That means, to get the graph of y, the red curve above `(y_1)` should be moved in vertical direction, which is 5 units down.

**Hence, the graph of the given function `y=(x+2)^3-5` is:**

`y = (x+2)^3-5`

When y = 0;

`0 = (x+2)^3-5`

`x = (5)^(1/3)-2 = -0.29`

When `x = 0` ;

`y = (0+2)^3-5`

`y = 3`

So the x intercept of the graph is `y = 3` and y intercept of the graph is `x = -0.29` .

The maximum and minimum for this graph is obtained when `y' = 0` .

`y' = 3(x+2)^2`

When `y' = 0` ;

`3(x+2)^2 = 0`

`x = -2`

`y'' = 6(x+2)`

`(y'')_(x=-2) = 0`

So we have a point of inflection at `x = -2` .

When `x = -2` then `y = -5`

When `x rarr +oo;`

`lim_(xrarroo)(x+2)^3-5 = +oo`

When `x rarr -oo;`

`lim_(xrarr-oo)(x+2)^3-5 = -oo`

*So using the above details we can plot the graph. *

*The plotted graph is shown below.*

**Sources:**

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