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3 years ago

Which equation is y= 2x ^2 - 8x +9 rewritten in vertex form.

1 answer:

7 0

y= 2x ^2 - 8x +9

y = a(x - h)2 + k, where (h, k) is the vertex of the parabola
so
y= 2x ^2 - 8x + 9
y= 2x ^2 - 8x + 8 + 1
y = 2(x^2 - 4x - 4) + 1
y = 2(x - 2)^2 + 1 ....<---------vertex form

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Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5),

larisa86 [58]

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$f(t)=-0.2+4.1sin(t)+4cos(t)$

Step-By-Step Explanation

Given the function $f(t)=c_0+c_1sin(t)+c_2cos(t)$ .

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

$f(0)=c_0+0c_1+c_2=5.5$ .

$f(\pi /2)=c_0+c_1+0c_2=0.5$ .

$f(\pi)=c_0+0sin(t)-c_2=-2.5$ .

$f(3\pi /2)=c_0-c_1+0c_2=-7.5$ .

Expressing this as a system of linear equations in matrix form AX=B

$\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right)$

Where

$A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)$ ,

$B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)$

To determine the values of X, we use the expression

$X=(A^{T}A)^{-1}A^{T}B$

$A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right)$

$(A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$

$A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

Therefore:

$X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)=\left(\begin{array}{c} -0.2 \\4.1\\4\end{array}\right)$

Therefore, the trigonometric function which fits to the given data is:

$f(t)=-0.2+4.1sin(t)+4cos(t)$

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3 years ago