Answer:
See proof below
Step-by-step explanation:
<u>Important points</u>
- understanding what it means to be "onto"
- the nature of a quadratic function
- finding a value that isn't in the range
<u>Onto</u>
For a function with a given co-domain to be "onto," every element of the co-domain must be an element of the range.
However, the co-domain here is suggested to be , whereas the range of f is not (proof below).
<u>Proof (contradiction)</u>
Suppose that f is onto .
Consider the output 7 (a specific element of ).
Since f is onto , there must exist some input from the domain , "p", such that f(p) = 7.
Substitute and solve to find values for "p".
Next, apply the square root property:
By definition, , so
By the Fundamental Theorem of Algebra, any polynomial of degree n with complex coefficients, has exactly n complex roots. Since the degree of f is 2, there are exactly 2 roots, and we've found them both, so we've found all of them.
However, neither nor are in , so there are zero values of p in for which f(p)= 7, which is a contradiction.
Therefore, the contradiction supposition must be false, proving that f is not onto