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Vladimir79 [104]
3 years ago
12

"The graph of F(x), shown below, resembles the graph of G(x) = x2, but it has been changed somewhat. Which of the following coul

d be the equation of F(x)?

Mathematics
2 answers:
SpyIntel [72]3 years ago
5 0
G(x)=x²
The graph  has moved to the right 4 units, therefore the new graph will be:
H(x)=(x-4)²

It has also move 4 units up, therefore the new graph will be:
F(x)=(x-4)²+4

Answer: 
F(x)=(x-4)²+4

WINSTONCH [101]3 years ago
5 0

y = (x - 4)² + 4

or y = x² - 8x + 20

<h3>Further explanation</h3>

Transformation of a graph is changing the shape and location of a graph.

There are four types of transformation geometry: translation (or shifting), reflection, rotation, and dilation (or stretching).  

  • In this case, the transformation is shifting horizontally or vertically.
  • Translation (or shifting): moving a graph on an analytic plane without changing its shape.
  • Vertical shift: moving a graph upwards or downwards without changing its shape.
  • Horizontal shift: moving a graph to the left or right downwards without changing its shape.  

In general, given the graph of y = f(x) and v > 0, we obtain the graph of:

  • \boxed{ \ y = f(x) + v \ } by shifting the graph of \boxed{ \ y = f(x) \ } upward v units.
  • \boxed{ \ y = f(x) - v \ } by shifting the graph of \boxed{ \ y = f(x) \ } downward v units.

That's the vertical shift, now the horizontal one. Given the graph of y = f(x) and h > 0, we obtain the graph of:

  • \boxed{ \ y = f(x + h) \ } by shifting the graph of \boxed{ \ y = f(x) \ } to the left h units.
  • \boxed{ \ y = f(x - h) \ } by shifting the graph of \boxed{ \ y = f(x) \ } to the right h units.

Hence, the combination of vertical and horizontal shifts is as follows:

\boxed{\boxed{ \ y = f(x \pm h) \pm v \ }}

The plus or minus sign follows the direction of the shift, i.e., up-down or left-right

<u>Given:</u> \boxed{ \ g(x) = x^2 \ becomes \ f(x) = ? \ }

In the graph, notice the shifting of the vertex from (0, 0) to (4, 4).

From this, we can describe that from g(x) to f(x) there has been a shift to the right 4 units and upward 4 units.

Let us construct f(x) from g(x).

\boxed{ \ g(x) = y = x^2 \ } \rightarrow \boxed{ \ f(x) = y = (x + h)^2 + v \ }

We set h = -4 and v = +4 and we get the equation f(x) as

\boxed{\boxed{ \ f(x) = (x - 4)^2 + 4 \ }}

Let's expand it if we want to represent a standard form of a quadratic function, like this:

\boxed{ \ f(x) = x^2 - 8x + 16 + 4 \ }

\boxed{\boxed{ \ f(x) = x^2 - 8x + 20 \ }}

<u>Conclusion </u>

The graph of f(x) is drawn by the combination of shifting the graph of g(x) to the right 4 units and upward 4 units.  

<h3>Learn more  </h3>
  1. Transformations that change the graph of (f)x to the graph of g(x) brainly.com/question/2415963
  2. The similar problem brainly.com/question/1369568
  3. Determine the coordinates of the image of a point after the triangle is rotated 270° about the origin brainly.com/question/7437053

Keywords: transformations, the graph of f(x), resembles, g(x) = x², f(x) = (x - 4)² + 4, y = x² - 8x + 20, translation, shifting, right, upward , horizontal, vertical

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6 0
2 years ago
46h+56h=600 please solve
valentinak56 [21]

Answer:

5.88235294118

Step-by-step explanation:

Step 1:

46h + 56h = 600       Equation

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Hope This Helps :)

3 0
3 years ago
What is 5.316 - 1.942 btw (show ur work) :)
erastovalidia [21]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4\\

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}\\

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&11&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&\linethrough{1}&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{3}&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

6 0
2 years ago
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