32. (a) For an even function, f(x) = f(-x). Given f(5) = 3, we know f(-5) = 3.
Therefore (-5, 3) is also on the graph.
For an odd function, f(-x) = -f(x). Given f(5) = 3, we know f(-5) = -3.
Therefore (-5, -3) is also on the graph.
33. f(-x) = -f(x). The function is odd.
34. f(-x) = x/(x-1) ≠ -f(x) ≠ f(x). The function is neither even nor odd.
35. f(-x) = f(x). The function is even.
A, B, and C. all of them are reasonable choices and wouldn’t give any conclusion, because not everyone has the flu
I solved it for you. hopefully this helps
Answer:
Step-by-step explanation:u have to add x+6 = ur answer there u go
Let's to the first example:
f(x) = x^2 + 9x + 20
Ussing the formula of basckara
a = 1
b = 9
c = 20
Delta = b^2 - 4ac
Delta = 9^2 - 4.(1).(20)
Delta = 81 - 80
Delta = 1
x = [ -b +/- √(Delta) ]/2a
Replacing the data:
x = [ -9 +/- √1 ]/2
x' = (-9 -1)/2 <=> - 5
Or
x" = (-9+1)/2 <=> - 4
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Already the second example:
f(x) = x^2 -4x -60
Ussing the formula of basckara again
a = 1
b = -4
c = -60
Delta = b^2 -4ac
Delta = (-4)^2 -4.(1).(-60)
Delta = 16 + 240
Delta = 256
Then, following:
x = [ -b +/- √(Delta)]/2a
Replacing the information
x = [ -(-4) +/- √256 ]/2
x = [ 4 +/- 16]/2
x' = (4-16)/2 <=> -6
Or
x" = (4+16)/2 <=> 10
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Now we are going to the 3 example
x^2 + 24 = 14x
Isolating 14x , but changing the sinal positive to negative
x^2 - 14x + 24 = 0
Now we can to apply the formula of basckara
a = 1
b = -14
c = 24
Delta = b^2 -4ac
Delta = (-14)^2 -4.(1).(24)
Delta = 196 - 96
Delta = 100
Then we stayed with:
x = [ -b +/- √Delta ]/2a
x = [ -(-14) +/- √100 ]/2
We wiil have two possibilities
x' = ( 14 -10)/2 <=> 2
Or
x" = (14 +10)/2 <=> 12
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To the last example will be the same thing.
f(x) = x^2 - x -72
a = 1
b = -1
c = -72
Delta = b^2 -4ac
Delta = (-1)^2 -4(1).(-72)
Delta = 1 + 288
Delta = 289
Then we are going to stay:
x = [ -b +/- √Delta]/2a
x = [ -(-1) +/- √289]/2
x = ( 1 +/- 17)/2
We will have two roots
That's :
x = (1 - 17)/2 <=> -8
Or
x = (1+17)/2 <=> 9
Well, this would be your answers.
