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] .
A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic.
Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation
of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15)
it's standard equation is x^2 + y^2/16 =1
tell my why is it only x^2? why is there no a^2?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.