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Mathematics

1 answer:

Sergeu [11.5K]2 years ago

5 0

Using translation concepts, it is found that the transformations to create function d are given as follows:

Horizontal shift right 1 unit.

Vertical shift up 5 units.

Frequency multiplied by 2.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem, the parent cosine function is given by:

f(x) = cos(x).

The translated function is given by:

d(x) = cos(2x - 1) + 5.

Which means that:

1 was subtracted in the domain, hence the was a horizontal shift right 1 unit.

5 was added in the range, hence there was a vertical shift up 5 units.

There was a multiplication by 2 in the domain, hence the frequency is multiplied by 2.

More can be learned about translation concepts at brainly.com/question/4521517

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For the polynomial function ƒ(x) = x4 − 12x3 + 46x2 − 60x + 25, find all local and global extrema.

rodikova [14]

$\bf f(x)=x^4-12x^3+46x^2-60x+25\\\\
-------------------------------\\\\
\cfrac{df}{dx}=4x^3-36x^2+92x-60\implies 0=4x^3-36x^2+92x-60
\\\\\\
0=x^3-9x^2+23x-15\impliedby \textit{doing some synthetic division with 5}
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0=(x^2-4x+3)(x-5)\implies 0=(x-3)(x-1)(x-5)\\\\
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\begin{cases}
f(1)=0\leftarrow minimum&1,0\\
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4 0

3 years ago

The graph of F(x) shown below has the same shape as the graph of G(x) = x^2 but it is shifted down 5 units and to the left 4 uni

grandymaker [24]

$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}$

$\bf \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's see

down by 5 units, D = -5

to the left by 4 units, C = +4

$\bf G(x)=x^2\implies G(x)=1(1x+\stackrel{C}{0})^2+\stackrel{D}{0} \\\\\\ \begin{cases} D=-5\\ C=+4 \end{cases}\implies F(x)=1(1x+\stackrel{C}{4})^2\stackrel{D}{-5}\implies F(x)=(x+4)^2-5$