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AveGali [126]
3 years ago
5

Which statements about the function described by the table is true

Mathematics
1 answer:
puteri [66]3 years ago
6 0

Answer: Can you give more detail

Step-by-step explanation:

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-3x to the fourth power -3x to the third power+x squared-2x to the third power -xsquared -8x +6
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Given the sum:

\begin{gathered} -3x^4-3x^3+x^2 \\ \ldots\ldots-2x^3-x^2-8x+6 \end{gathered}

The sum is given below:

Therefore, the sum of the polynomials is:

-3x^4-5x^3-8x+6

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Given cos(4x)+2<br><br> Find: Period, Shift, &amp; Range
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\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

\bf \begin{array}{llll}&#10;&#10;&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{function period or frequency}\\&#10;\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)&#10;\end{array}

now, with that template in mind, let's see

\bf \begin{array}{llccll}&#10;cos(&4x&+0)&+2\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&B&C&D&#10;\end{array} &#10;\\\\\\&#10;period\qquad \cfrac{2\pi }{B}\implies \cfrac{2\pi }{4}\implies \cfrac{\pi }{2}&#10;\\\\\\&#10;\textit{horizontal/phase shift}\qquad \cfrac{C}{B}\implies \cfrac{0}{4}\implies 0\impliedby none&#10;\\\\\\&#10;\textit{vertical shift}\qquad D=+2\impliedby \textit{2 units up}

now the range, how far up and down it goes on the y-axis

well, for the graph of cos(x), the range is, goes up to 1, down to -1, is all,

the midline is at 0

now, with a vertical shift of 2 upwards, it moves the midline by 2 units, used to be at 0, now is at y = 2, but the amplitude never changed, is goes up and down still one unit,

but with the midline at 2, goes up to 3 and down to 1, so the range is 3 ⩽ y ⩽ 1
4 0
4 years ago
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Aleks04 [339]

Answer:

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Step-by-step explanation:

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photoshop1234 [79]

Answer:

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Step-by-step explanation:

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