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9514 1404 393
Answer:
D: all real numbers
R: f(x) > 0
A: f(x) = 0
(-∞, 0), (+∞, +∞)
vertical stretch by a factor of 2; left shift 2 units
Step-by-step explanation:
The transformation ...
g(x) = a·f(b(x -c)) +d
does the following:
- vertical stretch by a factor of 'a'
- horizontal compression by a factor of 'b'
- translation right by 'c' units
- translation up by 'd' units
For many functions, horizontal coordinate changes are indistinguishable from vertical coordinate changes. Exponential functions tend to be one of those.
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Using the above notation, you seem to have f(x) = 3^x, and g(x) = 2f(x+2). The transformation is a vertical stretch by a factor of 2, and a translation left 2 units.
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As with all exponential functions, ...
- the domain is "all real numbers"
- the range is all numbers above the asymptote: f(x) > 0
- the horizontal asymptote is f(x) = 0
The function is a growth function, so ...
- x → -∞, f(x) → 0
- x → ∞, f(x) → ∞
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<em>Additional comment</em>
The left shift is equivalent to an additional vertical stretch. The function could be rewritten as ...
f(x) = 18(3^x)
with no left shift and a vertical stretch by a factor of 18 instead of 2.
Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
