SOMEONE PLEASE EXPLAIN ASAP
1 answer:
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Our function f(x) can be rewritten if we factor out a common x^2 from each term:
Now inside the parentheses we have a polynomial of the form a^2 - b^2, or the difference of two perfect squares, which can be factored as (a+b)(a-b) so we have:
Setting our function equal to zero gives us the roots x = 0, x = 4, and x = -4.
The multiplicity of the root zero is two since it occurs twice, and the others are one since they occur only once. If you graph the function you can see that it will only touch the x-axis at x = 0, but will cross the x-axis at x = 4 and x = -4.
Now inside the parentheses we have a polynomial of the form a^2 - b^2, or the difference of two perfect squares, which can be factored as (a+b)(a-b) so we have:
Setting our function equal to zero gives us the roots x = 0, x = 4, and x = -4.
The multiplicity of the root zero is two since it occurs twice, and the others are one since they occur only once. If you graph the function you can see that it will only touch the x-axis at x = 0, but will cross the x-axis at x = 4 and x = -4.
Answer:
The second choice, .
Step-by-step explanation:
Note, that the expression is an equation. A point
is on the graph of
if and only if the value of
and
satisfy this equation; that is: in other words, the
-coordinate of that point (the second number in the tuple) should be equal to
, which is equal to
(evaluated where
is equal to the first number in the tuple.
For each tuple in the choices, calculate the value of where
is equal to the first number of each tuple. Compare the result to the second number in that tuple. That choice corresponds to a valid point on
only if these two numbers match.
- First choice:
,
. That's not the same as the second number,
. Therefore, this point isn't on the graph of
.
- Second choice:
,
. That matches the second number in the tuple. Therefore, this point is on the graph of
- Third choice:
,
. That's not the same as the second number,
. Therefore, this point isn't on the graph of
- Fourth choice:
,
. That's not the same as the second number,
. Therefore, this point isn't on the graph of