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] .
The anwser is imaginary
number if you want to know The anwser of the expression
you must arrange The expression and use quadratic formula To find the
answer
master
The anwser is imaginary
number if you want to know The anwser of the expression
you must arrange The expression and use quadratic formula To find the
answer
master
Y
master
X2-2X+8-4X2+12X-20=0
(X2-4X2)+(-2X+12X)+(-20+8)= 0
-3X2+10X-12=0
3X2-10X+12=0
Use quadratic formula To find the answer
answer (5±Root11i)/3
master
Soo sorry (5±Root11* i)/3
master
x2-2x+8-4x2+12x-20
x2-4x2-2x+12x+8-20
-3x2+10x-12
now you can find the answer using quadratic
Mukhtar
explain and give four example of hyperbolic function
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.