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Shifts of the parent function f ( x ) = b x

For any constants c and d , the function f ( x ) = b x + c + d shifts the parent function f ( x ) = b x

  • vertically d units, in the same direction of the sign of d .
  • horizontally c units, in the opposite direction of the sign of c .
  • The y -intercept becomes ( 0 , b c + d ) .
  • The horizontal asymptote becomes y = d .
  • The range becomes ( d , ) .
  • The domain, ( , ) , remains unchanged.

Given an exponential function with the form f ( x ) = b x + c + d , graph the translation.

  1. Draw the horizontal asymptote y = d .
  2. Identify the shift as ( c , d ) . Shift the graph of f ( x ) = b x left c units if c is positive, and right c units if c is negative.
  3. Shift the graph of f ( x ) = b x up d units if d is positive, and down d units if d is negative.
  4. State the domain, ( , ) , the range, ( d , ) , and the horizontal asymptote y = d .

Graphing a shift of an exponential function

Graph f ( x ) = 2 x + 1 3. State the domain, range, and asymptote.

We have an exponential equation of the form f ( x ) = b x + c + d , with b = 2 , c = 1 , and d = 3.

Draw the horizontal asymptote y = d , so draw y = −3.

Identify the shift as ( c , d ) , so the shift is ( 1 , −3 ) .

Shift the graph of f ( x ) = b x left 1 units and down 3 units.

Graph of the function, f(x) = 2^(x+1)-3, with an asymptote at y=-3. Labeled points in the graph are (-1, -2), (0, -1), and (1, 1).

The domain is ( , ) ; the range is ( 3 , ) ; the horizontal asymptote is y = −3.

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Graph f ( x ) = 2 x 1 + 3. State domain, range, and asymptote.

The domain is ( , ) ; the range is ( 3 , ) ; the horizontal asymptote is y = 3.

Graph of the function, f(x) = 2^(x-1)+3, with an asymptote at y=3. Labeled points in the graph are (-1, 3.25), (0, 3.5), and (1, 4).
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Given an equation of the form f ( x ) = b x + c + d for 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 f ( x ) 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 f ( x ) .
  • To find the value of 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 42 = 1.2 ( 5 ) x + 2.8 graphically. Round to the nearest thousandth.

Press [Y=] and enter 1.2 ( 5 ) x + 2.8 next to Y 1 =. Then enter 42 next to Y2= . For a window, use the values –3 to 3 for x and –5 to 55 for y . Press [GRAPH] . The graphs should intersect somewhere near 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, x 2.166.

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Solve 4 = 7.85 ( 1.15 ) x 2.27 graphically. Round to the nearest thousandth.

x 1.608

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Graphing a stretch or compression

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 f ( x ) = b x by a constant | a | > 0. For example, if we begin by graphing the parent function f ( x ) = 2 x , we can then graph the stretch, using a = 3 , to get g ( x ) = 3 ( 2 ) x as shown on the left in [link] , and the compression, using a = 1 3 , to get h ( x ) = 1 3 ( 2 ) x as shown on the right in [link] .

Questions & Answers

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SLIMANE
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rachel Reply
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
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SLIMANE
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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?
Reena Reply
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Reena Reply
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
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Den Reply
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Jeffrey
Confunction Identity
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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.
Shakeena Reply
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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