Answer:
[tex]\textsf{a)\;\;passes\;through\;$(-2, 18.75)$\;and\;$(1, 1.2)$}[/tex]
[tex]\textsf{$a$-value:\;\;3 \quad $b$-value: \;\;0.4}[/tex]
[tex]\textsf{Equation: \quad $y=3 (0.4)^x$}[/tex]
[tex]\textsf{b)\;\;passes\;through\;$(0, 6)$\;and\;$(2, 8.64)$}[/tex]
[tex]\textsf{$a$-value:\;\;6 \quad $b$-value: \;\;1.2}[/tex]
[tex]\textsf{Equation: \quad $y=6 (1.2)^x$}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{9 cm}\underline{General form of an Exponential Function}\\\\$y=ab^x$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $b$ is the base (growth/decay factor) in decimal form.\\\end{minipage}}[/tex]
Part (a)
Given points:
Substitute both points into the exponential function formula to create two equations:
- [tex]\textsf{Equation\;1}: \quad 18.75=ab^{-2}[/tex]
- [tex]\textsf{Equation\;2}: \quad 1.2=ab[/tex]
Divide the equations to eliminate a, then solve for b:
[tex]\implies \dfrac{18.75}{1.2}=\dfrac{ab^{-2}}{ab}[/tex]
[tex]\implies15.625=\dfrac{b^{-2}}{b}[/tex]
[tex]\implies15.625=b^{-2}b^{-1}[/tex]
[tex]\implies15.625=b^{-3}[/tex]
[tex]\implies15.625=\dfrac{1}{b^{3}}[/tex]
[tex]\implies b^{3}=\dfrac{1}{15.625}[/tex]
[tex]\implies b=0.4[/tex]
Substitute the found value of b into the second equation and solve for b:
[tex]\implies 1.2=0.4a[/tex]
[tex]\implies a=3[/tex]
Therefore, the exponential equation is:
[tex]y=3 (0.4)^x[/tex]
-------------------------------------------------------------------------------------------
Part (b)
Given points:
Substitute point (0, 6) into the exponential function formula and solve for a:
[tex]\implies 6=ab^0[/tex]
[tex]\implies 6=a(1)[/tex]
[tex]\implies a=6[/tex]
Substitute the found value of a and point (2, 8.64) into the exponential function formula and solve for b:
[tex]\implies 8.64=6b^2[/tex]
[tex]\implies b^2=\dfrac{8.64}{6}[/tex]
[tex]\implies b^2=1.44[/tex]
[tex]\implies b=\sqrt{1.44}[/tex]
[tex]\implies b=1.2[/tex]
Therefore, the exponential equation is:
[tex]y=6 (1.2)^x[/tex]
