Question

Exponential function f(x) has a y intercept of 3 and an x intercept of -2. the function is always increasing as the value of x increases, but the function never reaches y=4
would it be decreasing or increasing and will it point down or up?
PLZ ANSWER ASAP

Answer 1

Answer:

$f(x)=-\left(\dfrac{1}{2}\right)^x+4$

The exponential function is increasing and is concave down

Step-by-step explanation:

Let the equation of exponential function be

$f(x)=a\cdot b^x+c,\ ,\ b>0$

1. Exponential function f(x) has a y intercept of 3, so the graph of f(x) passes through the point (0,3). Thus,

$3=a\cdot b^0+c\\ \\3=a+c$

2. Exponential function f(x) has an x intercept of -2, so the graph of f(x) passes through the point (-2,0). Thus,

$0=a\cdot b^{-2}+c\\ \\0=\dfrac{a}{b^2}+c$

3. Function is always increasing as the value of x increases, but the function never reaches y=4, so

$c=4$

Hence,

$\left\{\begin{array}{l}a+4=3\\ \\\dfrac{a}{b^2}+4=0\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=-1\\ \\\dfrac{-1}{b^2}+4=0\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=-1\\ \\b^2=\dfrac{1}{4}\end{array}\right.$

So,

$a=-1,\\ \\b=\dfrac{1}{2},\\ \\c=4,\\ \\f(x)=-\left(\dfrac{1}{2}\right)^x+4$