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SIZIF [17.4K]
3 years ago
14

How do I graph a function based off the unit circle?

Mathematics
1 answer:
sweet [91]3 years ago
7 0
So if you remember what the normal y = sin(x) function looks like (a wave), y = 2 sin(4x) is just changed a little.

The standard format for sine/cosine function
<span>y = a sin<span>(bx− c)</span> + <span>d

a = amplitude, distance from center of the wave to the highest point. This function a = 2 so the height of the sine wave reaches 2 instead of 1.
"c" and "d" shift the graph left/right and up/down respectively. These equal zero so the sine wave is not shifted.

The range (y-values) is then just the amplitude -2 ≤ y ≤ 2

The domain (x-value) is all real numbers because the wave just keeps going on to infinity in both directions.

2π / |b| = period, distance per wave
this equation b = 4
period is then π/2
this is the distance before a wave repeats.

Graph
     x  |  y
-π/8    -2
    0      0
  π/8    2
3π/8   -2
5π/8    2

see the pattern? I'm using the amplitude or peaks and bottoms of the wave y = 2 and -2  then using the x-distance between like points is the period so you add π/2

(π/8 , 2) 
+ π/2
(5π/8 , 2)

Same for the minumums of the wave (y = -2)
(-π/8 , -2)
+ π/2
(3π/8 , -2)

Hope this helps, otherwise there are youtube videos you can watch or try an online graphing calculator like Desmos.com





</span></span><span />
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aggreed very mean.....

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2 years ago
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Write a quadratic function with a vertex at (-9,13).
Lorico [155]

Answer:

y = x² + 18x + 94

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (- 9, 13) and letting a = 1, then

y = (x + 9)² + 13 ← expand (x + 9)² using FOIL

   = x² + 18x + 81 + 13

   = x² + 18x + 94

5 0
3 years ago
Read 2 more answers
18. Let f be a function with domain the set of all real numbers and having the following
Mashcka [7]

The derivative of the given function is f'(x) = k f(x) where k= \lim_{h \to 0} \frac{f(h)-1}{h}.

<h3>What is the derivative of a function?</h3>

Let f be a function defined on a neighborhood of a real number a. Then f is said to be differentiable or derivable at 'a' if \lim_{x \to a} \frac{f(x)-f(a)}{x-a} exists finitely. The limit is called the derivative or differential coefficient of f at 'a'. It is denoted by f'(a).

If f is differentiable at 'a', then

f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}

<h3>Calculation:</h3>

The given properties are:

(i) f(x + y) = f(x)f(y) for all real numbers x and y.

(ii) \lim_{h \to 0} \frac{f(h)-1}{h} = k; where k is a nonzero real number.

Then, the derivative of the function f(x) is,

f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

From property (i), f(x + h) = f(x)f(h)

On substituting,

f'(x) = \lim_{h \to 0} \frac{f(x)f(h)-f(x)}{h}

      = \lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}

From property (ii), \lim_{h \to 0} \frac{f(h)-1}{h} = k;

f'(x) = \lim_{h \to 0} \frac{f(x)[f(h) - 1]}{h}

      = f(x). \lim_{h \to 0} \frac{f(h)-1}{h}

      = f(x). k

      = kf(x)

Therefore, f'(x) = k f(x); where f'(x) exists for all real numbers of x.

Learn more about the derivative of a function here:

brainly.com/question/5313449

#SPJ9

6 0
2 years ago
Help me to answer this plss..
Hatshy [7]
Answer: 26ft

Divide 6.5 by .25 since .25 = 1/4
which equals 26
3 0
3 years ago
If x + 12 55- yand 5 - ys 2(x-3), then which statement is true?
Reil [10]

Answer:

C

Step-by-step explanation:

so

a≤b

b≤c

definitely a can't be more of c

at last, equal

so a≤c

7 0
3 years ago
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