MATH SOLVE

4 months ago

Q:
# Write an exponential function whose graph passes through the given points (2,8) (5,512)

Accepted Solution

A:

We are going to use the basic form of an exponential function: [tex]y=ab^x[/tex] to solve this problem.

Since we have tow points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex], we are going to have a system of equations:

[tex]y_{1}=ab^{x_{1}[/tex] equation (1)

[tex]y_{2}=ab^{x_{2}}[/tex] equation (2)

We know for our problem that [tex]x_{1}=2[/tex], [tex]y_{1}=8[/tex], [tex]x_{2}=5[/tex], and [tex]y_{2}=512[/tex], so lets replace those values in our equations:

[tex]8=ab^2[/tex] equation (1)

[tex]512=ab^5[/tex] equation (2)

Now, we just need to find [tex]a[/tex] and [tex]b[/tex]:

Dividing equation (2) by equation (1):

[tex] \frac{512=ab^5}{8=ab^2} [/tex]

[tex]64=b^3[/tex]

[tex]b= \sqrt[3]{64} [/tex]

[tex]b=4[/tex] equation (3)

Replacing equation (3) in equation (1):

[tex]8=a(4)^2[/tex]

[tex]8=16a[/tex]

[tex]a= \frac{8}{16} [/tex]

[tex]a= \frac{1}{2} [/tex]

Finally, we can put our function together:

[tex]y=ab^x[/tex]

[tex]y= \frac{1}{2} (4^x)[/tex]

We can conclude that the exponential function whose graph passes through the points (2,8) and (5,512) is [tex]y= \frac{1}{2} (4^x)[/tex].

Since we have tow points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex], we are going to have a system of equations:

[tex]y_{1}=ab^{x_{1}[/tex] equation (1)

[tex]y_{2}=ab^{x_{2}}[/tex] equation (2)

We know for our problem that [tex]x_{1}=2[/tex], [tex]y_{1}=8[/tex], [tex]x_{2}=5[/tex], and [tex]y_{2}=512[/tex], so lets replace those values in our equations:

[tex]8=ab^2[/tex] equation (1)

[tex]512=ab^5[/tex] equation (2)

Now, we just need to find [tex]a[/tex] and [tex]b[/tex]:

Dividing equation (2) by equation (1):

[tex] \frac{512=ab^5}{8=ab^2} [/tex]

[tex]64=b^3[/tex]

[tex]b= \sqrt[3]{64} [/tex]

[tex]b=4[/tex] equation (3)

Replacing equation (3) in equation (1):

[tex]8=a(4)^2[/tex]

[tex]8=16a[/tex]

[tex]a= \frac{8}{16} [/tex]

[tex]a= \frac{1}{2} [/tex]

Finally, we can put our function together:

[tex]y=ab^x[/tex]

[tex]y= \frac{1}{2} (4^x)[/tex]

We can conclude that the exponential function whose graph passes through the points (2,8) and (5,512) is [tex]y= \frac{1}{2} (4^x)[/tex].