How do you write an exponential equation that passes through (0, -2) and (2, -50)?

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Answer 1 · 2016-02-13T17:58:02

You set $f (x) = a \cdot e^{b x}$ $f(x)=a*e^(bx)$ and find the $a$ $a$ and $b$ $b$ constants via the 2 passing points.

$f (x) = - 2 \cdot e^{\frac{ln 25}{2} \cdot x}$ $f(x)=-2*e^(ln25/2*x)$
or
$f (x) = - 2 \cdot e^{1.60944 \cdot x}$ $f(x)=-2*e^(1.60944*x)$

Let the exponential function be:

$f (x) = a \cdot e^{b x}$ $f(x)=a*e^(bx)$

where $a$ $a$ and $b$ $b$ are constants to be found. From the two points that the function is passing we know that:

Point (0,-2)

$f (0) = - 2$ $f(0)=-2$

$a \cdot e^{b \cdot 0} = - 2$ $a*e^(b*0)=-2$

$a \cdot 1 = - 2$ $a*1=-2$

$a = - 2$ $a=-2$

Point (2,-50)

$f (2) = - 50$ $f(2)=-50$

$- 2 \cdot e^{b \cdot 2} = - 50$ $-2*e^(b*2)=-50$

$e^{2 b} = \frac{- 50}{- 2}$ $e^(2b)=(-50)/-2$

$e^{2 b} = 25$ $e^(2b)=25$

${ln e}^{2 b} = ln 25$ $lne^(2b)=ln25$

$2 b = ln 25$ $2b=ln25$

$b = \frac{ln 25}{2} = 1.60944$ $b=ln25/2=1.60944$

Function

So now that the constants are known:

$f (x) = - 2 \cdot e^{\frac{ln 25}{2} \cdot x}$ $f(x)=-2*e^(ln25/2*x)$
or
$f (x) = - 2 \cdot e^{1.60944 \cdot x}$ $f(x)=-2*e^(1.60944*x)$