For any constants
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}d,$ the function
$\text{\hspace{0.17em}}f(x)={b}^{x+c}+d\text{\hspace{0.17em}}$ shifts the parent function
$\text{\hspace{0.17em}}f(x)={b}^{x}$
vertically
$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units, in the
same direction of the sign of
$\text{\hspace{0.17em}}d.$
horizontally
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units, in the
opposite direction of the sign of
$\text{\hspace{0.17em}}c.$
The
y -intercept becomes
$\text{\hspace{0.17em}}\left(0,{b}^{c}+d\right).$
The horizontal asymptote becomes
$\text{\hspace{0.17em}}y=d.$
The range becomes
$\text{\hspace{0.17em}}\left(d,\infty \right).$
The domain,
$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ remains unchanged.
Given an exponential function with the form
$\text{\hspace{0.17em}}f(x)={b}^{x+c}+d,$ graph the translation.
Draw the horizontal asymptote
$\text{\hspace{0.17em}}y=d.$
Identify the shift as
$\text{\hspace{0.17em}}\left(-c,d\right).\text{\hspace{0.17em}}$ Shift the graph of
$\text{\hspace{0.17em}}f(x)={b}^{x}\text{\hspace{0.17em}}$ left
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is positive, and right
$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ units if
$c\text{\hspace{0.17em}}$ is negative.
Shift the graph of
$\text{\hspace{0.17em}}f(x)={b}^{x}\text{\hspace{0.17em}}$ up
$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units if
$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ is positive, and down
$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ units if
$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ is negative.
State the domain,
$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ the range,
$\text{\hspace{0.17em}}\left(d,\infty \right),$ and the horizontal asymptote
$\text{\hspace{0.17em}}y=d.$
Graphing a shift of an exponential function
Graph
$\text{\hspace{0.17em}}f(x)={2}^{x+1}-3.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.
We have an exponential equation of the form
$\text{\hspace{0.17em}}f(x)={b}^{x+c}+d,$ with
$\text{\hspace{0.17em}}b=2,$$\text{\hspace{0.17em}}c=1,$ and
$\text{\hspace{0.17em}}d=-3.$
Draw the horizontal asymptote
$\text{\hspace{0.17em}}y=d$ , so draw
$\text{\hspace{0.17em}}y=\mathrm{-3.}$
Identify the shift as
$\text{\hspace{0.17em}}\left(-c,d\right),$ so the shift is
$\text{\hspace{0.17em}}\left(-1,\mathrm{-3}\right).$
Shift the graph of
$\text{\hspace{0.17em}}f(x)={b}^{x}\text{\hspace{0.17em}}$ left 1 units and down 3 units.
The domain is
$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is
$\text{\hspace{0.17em}}\left(-3,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is
$\text{\hspace{0.17em}}y=\mathrm{-3.}$
Graph
$\text{\hspace{0.17em}}f(x)={2}^{x-1}+3.\text{\hspace{0.17em}}$ State domain, range, and asymptote.
The domain is
$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$ the range is
$\text{\hspace{0.17em}}\left(3,\infty \right);\text{\hspace{0.17em}}$ the horizontal asymptote is
$\text{\hspace{0.17em}}y=3.$
Given an equation of the form
$\text{\hspace{0.17em}}f(x)={b}^{x+c}+d\text{\hspace{0.17em}}$ for
$\text{\hspace{0.17em}}x,$ use a graphing calculator to approximate the solution.
Press
[Y=] . Enter the given exponential equation in the line headed “
Y
_{1} = ”.
Enter the given value for
$\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ in the line headed “
Y
_{2} = ”.
Press
[WINDOW] . Adjust the
y -axis so that it includes the value entered for “
Y
_{2} = ”.
Press
[GRAPH] to observe the graph of the exponential function along with the line for the specified value of
$\text{\hspace{0.17em}}f(x).$
To find the value of
$\text{\hspace{0.17em}}x,$ we compute the point of intersection. Press
[2ND] then
[CALC] . Select “intersect” and press
[ENTER] three times. The point of intersection gives the value of
x for the indicated value of the function.
Approximating the solution of an exponential equation
Solve
$\text{\hspace{0.17em}}42=1.2{\left(5\right)}^{x}+2.8\text{\hspace{0.17em}}$ graphically. Round to the nearest thousandth.
Press
[Y=] and enter
$\text{\hspace{0.17em}}1.2{\left(5\right)}^{x}+2.8\text{\hspace{0.17em}}$ next to
Y
_{1} =. Then enter 42 next to
Y2= . For a window, use the values –3 to 3 for
$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and –5 to 55 for
$\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Press
[GRAPH] . The graphs should intersect somewhere near
$\text{\hspace{0.17em}}x=2.$
For a better approximation, press
[2ND] then
[CALC] . Select
[5: intersect] and press
[ENTER] three times. The
x -coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for
Guess? ) To the nearest thousandth,
$\text{\hspace{0.17em}}x\approx \mathrm{2.166.}$
While horizontal and vertical shifts involve adding constants to the input or to the function itself, a
stretch or
compression occurs when we multiply the parent function
$\text{\hspace{0.17em}}f(x)={b}^{x}\text{\hspace{0.17em}}$ by a constant
$\text{\hspace{0.17em}}\left|a\right|>0.\text{\hspace{0.17em}}$ For example, if we begin by graphing the parent function
$\text{\hspace{0.17em}}f(x)={2}^{x},$ we can then graph the stretch, using
$\text{\hspace{0.17em}}a=3,$ to get
$\text{\hspace{0.17em}}g(x)=3{\left(2\right)}^{x}\text{\hspace{0.17em}}$ as shown on the left in
[link] , and the compression, using
$\text{\hspace{0.17em}}a=\frac{1}{3},$ to get
$\text{\hspace{0.17em}}h(x)=\frac{1}{3}{\left(2\right)}^{x}\text{\hspace{0.17em}}$ as shown on the right in
[link] .
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387