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] .
Questions & Answers
I've run into this:
x = r*cos(angle1 + angle2)
Which expands to:
x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2))
The r value confuses me here, because distributing it makes:
(r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1))
How does this make sense? Why does the r distribute once
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis
vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As
'f(x)=y'.
According to Google,
"The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks.
"Â" or 'Â' ... Â
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Period =2π
if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
the range is twice of the natural number which is the domain
Morolake
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?
For Plan A to reach $27/month to surpass Plan B's $26.50 monthly payment, you'll need 3,000 texts which will cost an additional $10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...