New Notes Blog: `(1,40) , (3, 640)` Write an exponential function `y=ab^x` whose graph passes through the given points.

Sunday, 13 July 2014

$\displaystyle{\left({1},{40}\right)},{\left({3},{640}\right)}$ 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 (1,40) and (3,640).

To determine the exponential function

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

plug-in the given x and y values.

For the first point (1,40), plug-in x=1 and y=40.

$\displaystyle{40}={a}{{b}}^{{1}}$

$\displaystyle{40}={a}{b}$ (Let this be EQ1.)

For the second point (3,640), plug-in x=3 and y=640.

$\displaystyle{640}={a}{{b}}^{{3}}$ (Let this be EQ2.)

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

The given two points of the exponential function are (1,40) and (3,640).

To determine the exponential function

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

plug-in the given x and y values.

For the first point (1,40), plug-in x=1 and y=40.

$\displaystyle{40}={a}{{b}}^{{1}}$

$\displaystyle{40}={a}{b}$ (Let this be EQ1.)

For the second point (3,640), plug-in x=3 and y=640.

$\displaystyle{640}={a}{{b}}^{{3}}$ (Let this be EQ2.)

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

$\displaystyle{40}={a}{b}$

$\displaystyle\frac{{40}}{{b}}={a}$

Plug-in this to EQ2.

$\displaystyle{640}={a}{{b}}^{{3}}$

$\displaystyle{640}={\left(\frac{{40}}{{b}}\right)}{{b}}^{{3}}$

And, solve for b.

$\displaystyle{640}={40}{{b}}^{{2}}$

$\displaystyle\frac{{640}}{{40}}={{b}}^{{2}}$

$\displaystyle{16}={{b}}^{{2}}$

$\displaystyle\pm\sqrt{{16}}={b}$

$\displaystyle\pm{4}={b}$

Take note that in 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 on the positive value of b which is 4.

Now that the value of b is known, plug-in it to EQ1.

$\displaystyle{40}={a}{b}$

$\displaystyle{40}={a}{\left({4}\right)}$

And, solve for a.

$\displaystyle\frac{{40}}{{4}}={a}$

$\displaystyle{10}={a}$

Then, plug-in the values of a and b to the exponential function

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

So this becomes:

$\displaystyle{y}={10}\cdot{{4}}^{{x}}$

Therefore, the exponential function that passes the given two points is $\displaystyle{y}={10}\cdot{{4}}^{{x}}$ .

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)