exponential and logarithmic functions graphing an exponential function and its asymptote: f(x) = a(e)^(x…

exponential and logarithmic functions graphing an exponential function and its asymptote: f(x) = a(e)^(x - b) - c graph the exponential function. g(x) = 2/3 e^(x + 4) - 1 plot two points on the graph of the function, and also draw the asymptote. then click on the graph - a - function button.

exponential and logarithmic functions graphing an exponential function and its asymptote: f(x) = a(e)^(x - b) - c graph the exponential function. g(x) = 2/3 e^(x + 4) - 1 plot two points on the graph of the function, and also draw the asymptote. then click on the graph - a - function button.

Answer

Explanation:

Step1: Encontrar el asintoto

Para una función exponencial de la forma $y = a e^{x - h}+k$, el asintoto horizontal es $y = k$. En la función $g(x)=\frac{2}{3}e^{x + 4}-1$, $k=-1$, así que el asintoto es $y = - 1$.

Step2: Encontrar un primer punto

Sustituimos $x=-4$ en la función $g(x)$. $g(-4)=\frac{2}{3}e^{-4 + 4}-1=\frac{2}{3}e^{0}-1=\frac{2}{3}\times1 - 1=\frac{2}{3}-1=-\frac{1}{3}$. Así, el primer punto es $(-4,-\frac{1}{3})$.

Step3: Encontrar un segundo punto

Sustituimos $x=-3$ en la función $g(x)$. $g(-3)=\frac{2}{3}e^{-3 + 4}-1=\frac{2}{3}e^{1}-1=\frac{2e}{3}-1\approx\frac{2\times2.718}{3}-1=\frac{5.436}{3}-1 = 1.812-1=0.812$. Así, el segundo punto es $(-3,0.812)$.

Answer:

Asintoto: $y = - 1$. Puntos: $(-4,-\frac{1}{3})$ y $(-3,0.812)$