Grade 11 Math Problem, cant figure out the transformations

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SUMMARY

The discussion focuses on the transformations of the exponential function y = 2^x, specifically how to derive the equation y = -2^(2x) + 6 from its mother graph. The user initially proposed y = -2^x + 6 but struggled to understand the transition to the exponent 2x. The correct approach involves recognizing that the graph's x-axis intersection point indicates a horizontal transformation, leading to the conclusion that the exponent must be 2x to achieve the desired transformation.

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  • Understanding of exponential functions and their graphs.
  • Familiarity with transformations of functions, including vertical and horizontal shifts.
  • Knowledge of logarithmic functions and their properties.
  • Ability to manipulate equations involving exponents.
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  • Learn how to apply logarithmic identities to solve exponential equations.
  • Explore graphical interpretations of function transformations.
  • Practice solving problems involving composite functions and their transformations.
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Students in Grade 11 mathematics, educators teaching exponential functions, and anyone looking to deepen their understanding of function transformations in algebra.

baller2353
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Homework Statement


the mother graph is y = 2 ^ x


Homework Equations


y=2 ^ x



The Attempt at a Solution


so i know the graph is flipped upsidedown and the whole graph is moved up 6 spots
so i can get y = -2 ^ x + 6

however the solutuion is y =-2 ^2x +6.
i can't seem to figure out how they got the exponent to the exponent 2x. I don't see what horizontal transformations could give you this.

question is in the picture attached.
thank u for your time.
 

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baller2353 said:

Homework Statement


the mother graph is y = 2 ^ x

Homework Equations


y=2 ^ x

The Attempt at a Solution


so i know the graph is flipped upsidedown and the whole graph is moved up 6 spots
so i can get y = -2 ^ x + 6

however the solutuion is y =-2 ^2x +6.
i can't seem to figure out how they got the exponent to the exponent 2x. I don't see what horizontal transformations could give you this.

question is in the picture attached.
thank u for your time.

Notice that for [tex]y=-2^x+6[/tex], it would cut the x-axis at [tex]x=\log_26=\frac{\ln6}{\ln2}\approx 2.58[/tex] but from the graph it looks like it's cutting the x-axis at about 1.2 or so, which is about half. Well, how do we get an answer of [tex]x=\frac{1}{2}\left(\log_26\right)[/tex] ? Solving backwards we have
[tex]2x=\log_26[/tex]
[tex]2^{2x}=6[/tex]
[tex]2^{2x}-6=0[/tex]

thus we needed an exponent of 2x. Now in an exam or even when solving other questions, you won't go into this much detail, all you need to do is look at where it approximately cuts the x-axis and then use a suitable integer scalar for the exponent (I doubt they'll ever use anything but integers, because it's just not detailed enough).
 

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