How do you write an exponential function whose graph passes through (0,-0.3) and (5,-9.6)?

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Answer 1 · 2016-11-03T17:06:39

Please see the explanation for steps leading to a solution.

An exponential function is:

$y = Ce^(alpha(x))$

We can find the value of C, using the point $(0, -0.3)$

$-0.3 = Ce^(alpha(0))$

$e^(alpha(0)) = 1$ so we merely flip the equation and drop the exponential:

$C = -0.3$

We can use the point $(5, -9.6)$ to find the value of $alpha$ :

$-9.6 = (-0.3)e^(alpha(5))$

Divide both side by $-0.3$

$(-9.6)/(-0.3) = e^(alpha(5))$

$32 = e^(alpha(5))$

Take the natural logarithm of both sides, to make the exponential disappear:

$ln(32) = alpha(5)$

Divide both sides by 5:

$alpha = ln(32)/5$

The above is the same as $ln(root(5)(32)) = ln(2)$

$alpha = ln(2)$

The exponential function is:

$y = (-0.3)e^(ln(2)x)$