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] .
Someone should please solve it for me
Add 2over ×+3 +y-4 over 5
simplify (×+a)with square root of two -×root 2 all over a
multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15
Second one, I got Root 2
Third one, I got 1/(y to the fourth power)
I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
graph the following linear equation using intercepts method.
2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b
you were already given the 'm' and 'b'.
so..
y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line.
where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2
2=3x
x=3/2
then .
y=3/2X-2
I think
Given
co ordinates for x
x=0,(-2,0)
x=1,(1,1)
x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
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
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
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.