Question

please help thx in adv

Answer 1

Answer:

Given function:

$\implies f(x)=\dfrac{100}{1+80e^{-0.3x}}$

y-intercept is when x = 0:

$\implies f(0)=\dfrac{100}{1+80e^{-0.3\cdot0}}$

$\implies f(0)=\dfrac{100}{81}=1.23 \ \sf(2 \ dp)$

Asymptotes:

y = 0

y = 100

Therefore, upper asymptote is y = 100

The function is in the form of the logistic growth model:

$f(x)=\dfrac{c}{1+ae^{-bx}}$

The point of maximum growth is $\left(\dfrac{\ln(a)}{b},\dfrac{c}{2} \right)$

Given:

• a = 80
• b = 0.3
• c = 100

Therefore, point of maximum growth is:

$\implies \left(\dfrac{\ln(80)}{0.3},\dfrac{100}{2} \right)$

$\implies (14.61, 50)$

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f(x) = 4 | x | - 1

y-intercept (0, -1)

x-intercepts:  (-0.25, 0)  (0.25, 0)

$f(x)=0.5 \lceil x \rceil+3$