How do you graph $f (x) = 1 + cos x$ $f(x)= 1+ cosx$ ?

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Answer 1 · 2018-04-23T06:21:38

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Please read the explanation.

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To graph $f (x) = 1 + cos (x)$ $color(red)(f(x)=1+Cos(x)$ ,

start working on it's Parent Function $f (x) = cos (x)$ $color(blue)(f(x) = Cos(x)$ first.

Make a table of values for $f (x) = cos (x) and f (x) = 1 + cos (x)$ $f(x)=Cos(x) and f(x)=1+cos(x)$

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For $x$ $color(red)(x$ , consider the values $0, \frac{π}{2}, π, \frac{3 π}{2} and 2 π$ $color(red)(0, pi/2, pi, (3pi)/2 and 2pi$ .

If you examine $Col 4 and Col 5$ $color(green)("Col 4" and "Col 5"$ , you see that the difference is 1.

$Graph of y = f(x) = cos(x)$ $color(red)("Graph of y = f(x) = Cos(x)"$

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$Graph of y = f(x) = 1 + cos(x)$ $color(red)("Graph of y = f(x) = 1 + Cos(x)"$

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$Graph of y = f(x) = cos(x)$ $color(red)("Graph of y = f(x) = Cos(x)")$ & $y = f(x) = 1 + cos(x)$ $color(blue)("y = f(x) = 1 + Cos(x)"$

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Use $y = A \cdot sin (B x + C) + D$ $color(green)(y=A*sin(Bx+C)+D$ [ or ]

use $y = A \cdot cos (B x + C) + D$ $color(green)(y=A*cos(Bx+C)+D$ ,

where, **Amplitude is ** $| A |$ $|A|$

**Period is ** $\frac{2 π}{B}$ $(2pi)/B$ and

Vertical Shift is $D$ $D$

Since $D = 1$ $D=1$ , the graph is shifted vertically by 1 unit.

If $D$ $D$ is given, the value of $D$ $D$ is responsible for a vertical shift.

Hope it helps.

How do you graph f(x)= 1+ cosx f(x)=1+cosxf(x)= 1+ cosx ?