Question

Which of the following graphs could be the graph of the function


`f(x)=0.03x^2(x^2-25)`

Answer 1

The answer to this problem which is the correct graph is the "FIRST graph". We have the given function:

f(x) = 0.03x² (x²-25)

From our evaluation, the has a power of "4".
When we try to start plotting points, we can check if the is reflecting these points.

First point is when x=1,
y=0.03x²(x²-24)
y =-0.72

The second point is when x= 6
y=0.03*6² *(6²-24)
y= 12.96

Only the FIRST graph satisfies these points.

Answer 2

The given function is

`f(x) = 0.03x^2 (x^2-25)`

To check out the graph, first we have to find the zeroes and there multiplicities .

TO find the zero, we put 0 for f(x), and solve for x, that is

`0.03x^2(x-5)(x+5)=0 \\ 0.03x^2 =0, x-5=0, x+5=0 \\ x=0 \ multiplicity \ 2, x=5 \ multiplicity \ 1, x=-5 \ multiplicity \ 1`

With odd multiplicity, the graph crosses the x axis and for even multiplicity, the graph takes a u turn at the zero .

So the correct graph is represented by first option .