Do all exponential functions go through the point (0, 1)?
If so, why is this the case?
The math book does not share the "WHY" with its readers.
I am just curious.
Thanks
Do all exponential functions go through the point (0, 1)?
If so, why is this the case?
The math book does not share the "WHY" with its readers.
I am just curious.
Thanks
Hello,
It depends on what you call an exponential function.
$\displaystyle f(x)=e^{ax}$, where a is a constant.
$\displaystyle f(0)=e^{a \cdot 0}=1$, so it goes through the point (0,1).
If $\displaystyle f(x)=be^{ax}$, where a is a constant and b is a non-zero constant, then $\displaystyle f(0)=be^{a \cdot 0}=b$, so it goes through the point (0,b).
Consider a simple exponential function with the base b. b must be positive and mostly b must be unequal 1. (With your question this constraint isn't necessary)
The equation of the function could be: $\displaystyle y = b^x$. A point of the graph of this function has the coordinates P(x, y) or in words:
P is on the graph if the coordinates consist of (exponent, value of the power). The base is considered to be a constant.
If the exponent is zero, the value of the power is $\displaystyle b^0$ which is by definition 1.
And therefore the graphs of all exponential functions with the equation mentioned above will pass through the point P(1, 0)
I don't want to sound rude, but what Moo used was not advanced math language. Someone in precalc should be able to understand what she said.
All she said was this:
It depends on how you define the exponential function.
If we have an exponential function $\displaystyle f(x)=a^x$ where $\displaystyle a$ has a constant value, the value of the function at x=0 is $\displaystyle f(0)=a^0=1$, thus it passes through the point $\displaystyle (0,1)$.
However, if you defind an exponential function as $\displaystyle f(x)=b\cdot a^x$, where b is some number, then the value of the function at x=0 is $\displaystyle f(0)=b\cdot a^0=b\cdot 1=b$, thus it passes through the point $\displaystyle (0,b)$
This shouldn't be that hard to grasp.
--Chris