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Artyom0805 [142]
2 years ago
7

What’s the end behavior of g(x)=(5^x-3)+2.5

Mathematics
1 answer:
fomenos2 years ago
3 0

Hello from MrBillDoesMath!

Answer:   g(x) approaches "infinity' as x approaches "positive infinity"  and g(x)  approaches --0.5 as x approaches "negative infinity"

Discussion.

g(x) = ( 5^x -3) + 2.5

Suppose x is approaching positive infinity. Then the term 5^x grows without bound  ("positive infinity")  so the -3 and "2.5" constants have no impact at all.  That is, g(x) behaves like 5^x which grows without bound.


Suppose x is approaching negative infinity. Then 5^x approaches  0 (zero) so g(x) approaches


(0 -3) + 2.5  -

= - 0.5


Thank you,

MrB

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\dfrac{\frac{2x}{3}-\frac{1}{2}}{\frac{3x}{4}-\frac{1}{2}} = \dfrac{\frac{4x-3}{6}}{\frac{3x-2}{4}} = \dfrac{4x-3}{6}\cdot \dfrac{4}{3x-2} = \dfrac{2(4x-3)}{3(3x-2)} = \dfrac{8x-6}{9x-6}

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36x-24x = 24-18 \iff 12x=6 \iff x = \dfrac{1}{2}

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