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zvonat [6]
1 year ago
7

Does anyone know how to graph trigonometric functions for algebra 2?

Mathematics
1 answer:
anastassius [24]1 year ago
4 0

Solution:

Give the following below

\begin{gathered} amplitude=3 \\ period=\frac{\pi}{2} \\ midline,\text{ }y=-1 \\ Passing\text{ through the point }(0,2) \end{gathered}

To find the cosine function, we will apply the general formula for cosine function below

\begin{gathered} f(x)=a\cos(bx-c)+d \\ Where \\ a\text{ is the amplitude} \\ b\text{ represents the speed of the cycle} \\ d\text{ is the vertical shift} \\ \frac{c}{b}\text{ is the phase shift \lparen horizontal shift\rparen} \\ Period=\frac{2\pi}{b} \end{gathered}

To find the value of b

\begin{gathered} Period=\frac{\pi}{2} \\ \frac{2\pi}{b}=\frac{\pi}{2} \\ Crossmultiply \\ b\times\pi=2\times2\pi \\ b=\frac{4\pi}{\pi} \\ b=4 \end{gathered}\begin{gathered} a=3 \\ b=4 \\ d=-1 \\ c=0 \end{gathered}

Substitute the values of the variables into the general formula for cosine function

\begin{gathered} f(x)=a\cos(bx-c)+d \\ f(x)=3\cos(4x-0)-1 \\ f(x)=3\cos4x-1 \end{gathered}

Applying a graphing tool,

The graph of one cycle of the function is shown below

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<span>The answer is AAS. </span><span>Angle Angle Side </span>postulate<span> </span><span>states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.</span>
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3 years ago
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A triangle has vertices at A ( 3 , 4 ) , B ( − 3 , 2 ) , C ( − 1 , − 4 ) . Is the triangle a right triangle? Explain.
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Answer: No

Step-by-step explanation:

The answer to this question is quite simple in the sense that all you need to do is graph the points on a graph and you can easily see that it is not a right triangle.

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3 years ago
If f(x)=3-2x and g(x)=1/x+5, What is the value of (f/g) (8)?
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(f/g)(8)=f(8)/g(8)

find f(8) and g(8) to mke it easier

f(8)=3-(8)=3-2(8)=3-16=-13

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so (f/g)(8)=-13/(1/13)=-13*13=-169

answer is -169

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4 years ago
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In two or more complete sentences, compare the number of x-intercepts in the graph of f(x)=x2 to the number of x-
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f(x) has 1 x intercept at x=0 because there is an x value that exists when y=0. For g(x), by adding 2 to x^{2} it moves the parent function up 2 so there is now no x-intercept.

Step-by-step explanation:

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4 years ago
Write a quadratic function in standard form with zeros 1 and -10
Licemer1 [7]

Answer:

f(x)=x^2+9x-10

Step-by-step explanation:

<u>Standard Form of Quadratic Function</u>

The standard form of a quadratic function is:

f(x)=ax^2+bx+c

where a,b, and c are constants.

The factored form of a quadratic equation is:

f(x)=a(x-\alpha)(x-\beta)

Where \alpha and \beta are the roots or zeros of f, and a is constant.

We know the zeros of the function are 1 and -10. The function is:

f(x)=a(x-1)(x-(-10))

f(x)=a(x-1)(x+10)

Operating:

f(x)=a(x^2+10x-x-10)

Joining like terms:

f(x)=a(x^2+9x-10)

Since we are not given any more restrictions, we can choose the value of a=1, thus. the required function is:

\boxed{f(x)=x^2+9x-10}

6 0
3 years ago
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