Question
Asked Oct 26, 2019

For the problem: f(x)=x^2+x-12/x+3

a. Find the x and y intercepts.

b. Please test for symmetry: is the function even, odd, or neither?

c. Find the vertical asymptotes

Thank you

check_circleExpert Solution
Step 1

To find x intercept set y=0 and solve for x.

Answer: x intercepts = -4 , 3

x2+x-12
f(x)=
X+3
x2
у3
2
X+3
x2+x-12
0=
X+4=0
X+3
x=-4
x2+x-12-0
х?+4x-3х-12-0
x(x+4)-3(x+4) 0
(X+4)(x-3)-0
X-3=0
x=3
help_outline

Image Transcriptionclose

x2+x-12 f(x)= X+3 x2 у3 2 X+3 x2+x-12 0= X+4=0 X+3 x=-4 x2+x-12-0 х?+4x-3х-12-0 x(x+4)-3(x+4) 0 (X+4)(x-3)-0 X-3=0 x=3

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Step 2

To find y intercept plug x=0 and solve for y.

Answer: y intercept = -4

x2+x-12
у3
х+3
02+0-12
у-
0+3
y-4
help_outline

Image Transcriptionclose

x2+x-12 у3 х+3 02+0-12 у- 0+3 y-4

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Step 3

Since, f(-x) is neither equal to f(x) or -f(x).

So it is neither e...

(-x)2+(-x)-12
f(-x)
(-x)+3
x2-x-12
f(-x)
X+3
x2+x-12
-f (x)=-
X+3
f(-x)f(x)
f(-x)-f(x)
help_outline

Image Transcriptionclose

(-x)2+(-x)-12 f(-x) (-x)+3 x2-x-12 f(-x) X+3 x2+x-12 -f (x)=- X+3 f(-x)f(x) f(-x)-f(x)

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