Q.Respected Sir/Madam; Sketch the graph of `y=(x+2)^3 - 5`

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jeew-m's profile pic

Posted on

`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.

lemjay's profile pic

Posted on


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:

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