New Notes Blog: `(1,2) , (3,50)` Write an exponential function `y=ab^x` whose graph passes through the given points.

Saturday, 24 January 2015

$\displaystyle{\left({1},{2}\right)},{\left({3},{50}\right)}$ Write an exponential function $\displaystyle{y}={a}{{b}}^{{x}}$ whose graph passes through the given points.

The given two points of the exponential function are (1,2) and (3,50).

To determine the exponential function

$\displaystyle{y}={a}{{b}}^{{x}}$

plug-in the given x and y values.

For the first point (1,2), plug-in x=1 and y=2.

$\displaystyle{2}={a}{{b}}^{{1}}$

$\displaystyle{2}={a}{b}$ (Let this be EQ1.)

For the second point (3,50), plug-in x=3 and y=50.

$\displaystyle{50}={a}{{b}}^{{3}}$ (Let this be EQ2.)

To solve for the values of a and b, apply substitution method of system of...

The given two points of the exponential function are (1,2) and (3,50).

To determine the exponential function

$\displaystyle{y}={a}{{b}}^{{x}}$

plug-in the given x and y values.

For the first point (1,2), plug-in x=1 and y=2.

$\displaystyle{2}={a}{{b}}^{{1}}$

$\displaystyle{2}={a}{b}$ (Let this be EQ1.)

For the second point (3,50), plug-in x=3 and y=50.

$\displaystyle{50}={a}{{b}}^{{3}}$ (Let this be EQ2.)

To solve for the values of a and b, apply substitution method of system of equations. To do so, isolate the a in EQ1.

$\displaystyle{2}={a}{b}$

$\displaystyle\frac{{2}}{{b}}={a}$

Plug-in this to EQ2.

$\displaystyle{50}={a}{{b}}^{{3}}$

$\displaystyle{50}={\left(\frac{{2}}{{b}}\right)}{{b}}^{{3}}$

And solve for b.

$\displaystyle{50}={2}{{b}}^{{2}}$

$\displaystyle\frac{{50}}{{2}}={{b}}^{{2}}$

$\displaystyle{25}={{b}}^{{2}}$

$\displaystyle\pm\sqrt{{25}}={b}$

$\displaystyle\pm{5}={b}$

Take note that in the exponential function $\displaystyle{y}={a}{{b}}^{{x}}$ , the b should be greater than zero $\displaystyle{\left({b}\gt{0}\right)}$ . When $\displaystyle{b}\leq{0}$ , it is no longer an exponential function.

So consider only the positive value of b which is 5.

Then, plug-in b=5 to EQ1.

$\displaystyle{2}={a}{b}$

$\displaystyle{2}={a}{\left({5}\right)}$

Isolate the a.

$\displaystyle\frac{{2}}{{5}}={a}$

Then, plug-in $\displaystyle{a}=\frac{{2}}{{5}}$ and $\displaystyle{b}={5}$ to

$\displaystyle{y}={a}{{b}}^{{x}}$

So this becomes:

$\displaystyle{y}=\frac{{2}}{{5}}\cdot{{5}}^{{x}}$

Therefore, the exponential function that passes the given two points is $\displaystyle{y}=\frac{{2}}{{5}}\cdot{{5}}^{{x}}$ .

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