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Question 189022This question is from textbook Introductory and Intermediate Algebra
: Hi, i tried solving my assignments, please help me check if i did it right..Thanks in advance...
1 Find the indicated function values:
f(x) =3x^2 + 4x - 2
a) f(0) d) f(b)
b) f(3) e) f(5a)
c) f(-5)
Solution:
a) f(0) = 3(0)^2 + 4(0) - 2
= - 2
b) f(3) = 3(3)^2 + 4(3) - 2
= 27 + 12 - 2
= 37
c) f(-5) = 3(-5)^2 + 4(-5) - 2
= 75 -20 - 2
= 53
d) f(b) = 3(b)^2 + 4(b) - 2
= 3b^2 + 4b - 2
e) f(5a) = 3(5a)^2 + 4(5a) - 2
= 75a^2 + 20a - 2
2 For f(x)= x^2 + 4x and g(x) = 2 - x
Find the following:
(f-g)(x) and (f-g)(6)
Solution:
(f-g) (x) = f(x) - g(x)
= ( x^2 + 4x ) - (2-x)
= x^2 + 4x - 2 + x
= x^2 + 5x - 2
Using (f-g)(x) = x^2 + 5x - 2
(f-g)(6) = (6)^2 + 5(6) -2
= 36 + 30 - 2
= 64
3 Find an equation of each line.Write the equation using function notation.Through (-4,8); perpendicular to 2x-3y = 1
Solution:
y = mx + b
3y = 2x - 1
y= (2x-1)/3
y = (2/3)x-3
y - y1 = m (x - x1)
(y-8) = 3/2 (x - (-4))
2(y-8) = 3(x+4)
2y - 16 = 3x + 12
2y = 3x + 28
y = (3x + 28)/2
f(x) = 3/2 x + 14
Regards,
Kattie
This question is from textbook Introductory and Intermediate Algebra
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 3 Find an equation of each line.Write the equation using function notation.Through (-4,8); perpendicular to 2x-3y = 1
Solution:
y = mx + b
3y = 2x - 1
y= (2x-1)/3
y = (2/3)x-3
y - y1 = m (x - x1)
(y-8) = 3/2 (x - (-4))
2(y-8) = 3(x+4)
2y - 16 = 3x + 12
2y = 3x + 28
y = (3x + 28)/2
f(x) = 3/2 x + 14
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The slope should be the negative inverse, -3/2 for perpendicular.
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