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garik1379 [7]
2 years ago
13

Suppose f (x) = x^2. Find the graph of f(x+2)

Mathematics
1 answer:
NISA [10]2 years ago
8 0
\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
~~~~y={{  A}}({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)={{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\\\
f(x)={{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\\\\
--------------------

\bf \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}
\\\\
\bullet \textit{ flips it sideways if }{{  B}}\textit{ is negative}\\
~~~~~~\textit{reflection over the y-axis}

\bf \bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
~~~~~~if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\left. \qquad  \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,

\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}

C = 2         B = 1        C/B = 2/1 or +2,    horizontal left shift of 2 units

f(x) shifted left by 2 units is f(x+2).
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Step-by-step explanation:

The question is as following:

The verticies of a triangle on the coordinate plane are

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=============================================

Given: the vertices of a triangle ABC are A(0, 0), B(2, 0) and C(0, 2).

IF the triangle is dilated by a factor of k about the origin, then

(x,y) → (kx , ky)

that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

It is given that triangle ABC was dilated by a factor of 1/3 to create the triangle A'B'C'.

If a figure dilated by a factor of 1/3 about the origin

So, (x,y)\rightarrow (\frac{1}{3}x,\frac{1}{3}y)

<u>So, The coordinates of the triangle A'B'C' are:</u>

A(0,0)\rightarrow A'(0,0) \\ B(2,0)\rightarrow B'(\frac{2}{3},0) \\ C(0,2)\rightarrow C'(0,\frac{2}{3})

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