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Sati [7]
1 year ago
8

Write a cosine function that Has a midline of 2 an amplitude of 3 and a period of 7pi/4

Mathematics
1 answer:
Alexxx [7]1 year ago
5 0

Given:

Amplitude of cosine function, A=3.

Period, T=7π/4.

Midline, D=2.

The time period can be expressed as:

T=\frac{2\pi}{B}

Put T=7π/4 to find the value of B.

\begin{gathered} \frac{7\pi}{4}=\frac{2\pi}{B} \\ B=\frac{4\times2}{7} \\ =\frac{8}{7} \end{gathered}

The general cosine function can be expressed as,

f(x)=A\cos (Bx)+D

Substitute B=8/7, A=3 and D=2 in above equation.

f(x)=3\cos (\frac{8}{7}x)+2

Therefore, the cosine function is,

f(x)=3\cos (\frac{8}{7}x)+2

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The minimum/maximum value of the function y = a(x − 2)(x − 1) occurs at x = d, what is the value of d?
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Answer:

d=\frac{3}{2}=1.5

Step-by-step explanation:

We have the function:

y=a(x-2)(x-1)

And we want to find x=d for which the minimum/maximum value will occur.

Notice that our function is a quadratic in factored form.

Remember that the minimum/maximum value always occurs at the vertex point.

And remember that the x-coordinate of the vertex is the axis of symmetry.

Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.

Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.

Our factors are <em>(x-2) </em>and <em>(x-1)</em>.

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d=\frac{1+2}{2}=3/2=1.5

Therefore, regardless of the value of <em>a</em>, the minimum/maximum value will occur at <em>x=d=1.5</em>.

Alternative Method:

Of course, we can also expand to confirm our answer. So:

y=a(x^2-2x-x+2)\\y=a(x^2-3x+2)\\y=ax^2-3ax+2a

The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.

And the formula for the vertex is:

x=-\frac{b}{2a}

So, we will substitute <em>-3a</em> for <em>b</em> and <em>a</em> for <em>a</em>. This yields:

x=-\frac{-3a}{2a}=\frac{3}{2}=1.5

Confirming our answer.

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