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saw5 [17]
3 years ago
14

Graph y = (x - 1) 2 + 2. Click on the graph until the graph of y = (x - 1) 2 + 2 appears.

Mathematics
2 answers:
strojnjashka [21]3 years ago
6 0
The vertex at (1,2) and from there use this to make the graph.

over 1, up 1. over 2 up 4, over 3 up 9. and if you know its right side, you know its left side too.
ehidna [41]3 years ago
3 0

Answer with explanation:

The given function is

  →   y= (x-1)²+2

→ y-2= (x-1)²

To find the vertex,

→y-2=0

→ y=2

And, x-1=0→x=1

So,the vertex of the given parabola is , (1,2).

To plot, the graph of the function find few ordered pair (x,y) and join them.

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Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.
Hitman42 [59]
\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
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sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
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sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
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\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}

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