Respuesta :

loneguy

[tex]\text{Given that,}\\\\f(x) = ab^x, f(0) = -5 ~ \text{And}~ f(3) = - \dfrac 5{27}\\\\\\\text{Now,}\\\\\\f(0) = -5 \\\\\implies ab^0 = -5\\\\\implies a =-5\\\\\\f(3) = -\dfrac{5}{27}\\\\\implies ab^3 = -\dfrac{5}{27}\\\\\implies -5b^3 = -\dfrac{5}{27}\\\\\implies b^3 = \dfrac 1{27}\\\\\\\implies b = \sqrt[3]{\dfrac{1}{27}} = \dfrac 13\\\\\\\text{Hence , } f(x) = ab^x = -5\left(\dfrac 13 \right)^x[/tex]

Answer:

f(x) = - 5 [tex](\frac{1}{3}) ^{x}[/tex]

Step-by-step explanation:

Using the conditions (0, - 5 ) and (3, - [tex]\frac{5}{27}[/tex] )

y = a[tex]b^{x}[/tex] ← using (0, - 5 )

- 5 = a[tex]b^{0}[/tex] [ [tex]b^{0}[/tex] = 1 ] , then

a = - 5

y = - 5[tex]b^{x}[/tex] ← using (3, - [tex]\frac{5}{27}[/tex] )

- [tex]\frac{5}{27}[/tex] = - 5b³ ( divide both sides by - 5 )

[tex]\frac{1}{27}[/tex] = b³ ( take cube root of both sides )

b = [tex]\sqrt[3]{\frac{1}{27} }[/tex] = [tex]\frac{1}{3}[/tex]

Then

f(x) = - 5 [tex](\frac{1}{3}) ^{x}[/tex]

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