Question

HELP FOR A MEDAL , NOT SURE IF IM CORRECT :/
To open these doors, you must speak three functions in standard form.
One function, f(x), with two real rational solutions.
f(x)=x2-4
One function, g(x), with two real irrational solutions.
g(x)= x2-2
One function, h(x), with two complex solutions.
h(x)= -3x2+11i
Create these three functions and explain to Professor McMerlock how you know these functions meet each condition. Remember, he is a Professor so use complete sentences. (Hint: Make sure that the b is even on g(x).)

Answer 1

Only few things you need to look at:
1. Your 'b' should be even on g(x). Standard form looks like this ax^2+bx+c your b=0 which is not an even number.
2. The h(x) function that you gave doesn't have a solution. It should look like h(x)=ax^2+b with 'a' and 'b' being positive.

Answer 2

@math&ing001 0 is an even number, as it is exactly divisible by 2.

Answer 3

omg ok do you think you can help me out? :/ im really confused and i need to turn this in tomorrow

Answer 4

One function, f(x), with two real rational solutions.
f(x)=x^2-4

that's correct, two solutions are x=2 and x=-2

One function, g(x), with two real irrational solutions.
g(x)= x^2-2

correct, solutions are x= sqrt 2 and x = -sqrt 2

One function, h(x), with two complex solutions.
h(x)= -3x2+11i

This doesn't have real coefficients... just make something that never touches the x axis,
like f(x) = x^2 + 1

Answer 5

Ok thank you! & can you explain to me how to convert the first equation into the general, vertex form ?

Answer 6

The first is in vertex form. You could write it as \[\large f(x) = (x-0)^2 - 4\]but that's the same thing. All three are in vertex form.

Answer 7

@agent0smith yeah it's getting late I should probably get some sleep
@omgkelley sry your second solution is correct and get the third function agent0 gave you.
Good luck

Answer 8

You were prob just thinking of positive numbers (it's neither pos or neg.)

Answer 9

ok thanks guys! :)