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AleksAgata [21]
3 years ago
12

Which translation maps the graph of the function f(x) = x2 onto the function g(x) = x2 + 2x + 6?

Mathematics
2 answers:
Flura [38]3 years ago
4 0
The graph went left 1 and up 6. You could complete the square to find out or just graph it!
Hope this helps!
Shtirlitz [24]3 years ago
4 0

Answer:

Translation: Shift 1 unit left and 5 unit up

Step-by-step explanation:

Given: f(x)=x^2

Translation function, g(x)=x^2+2x+6

First we change g(x) into vertex form and then check the translation.

g(x)=x^2+2x+6

Completing square

g(x)=(x^2+2x+1)+5

g(x)=(x+1)^2+5                     \because a^2+2ab+b^2=(a+b)^2

Now, we will compare the function with f(x)

f(x)=x^2

Shift f(x) 1 unit left

f(x)=(x+1)^2

Shift f(x) 5 unit up

f(x)=(x+1)^2+5=g(x)

Translation: 1 unit left and 5 unit up

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I need help with #11
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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}

now, with that template in mind, let's take a peek at this function

\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically
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