Q.Respected Sir/Madam; Sketch the graph of `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
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.