Y= x^2-4x+3 in vertex form

Mathematics

1 answer:

Karolina [17]2 years ago

4 0

Answer:

$y=(x-2)^2-1$

Step-by-step explanation:

<u>Complete the square</u>

<u /> $y=x^2-4x+3\\\\y+1=x^2-4x+3+1\\\\y+1=x^2-4x+4\\\\y+1=(x-2)^2\\\\y=(x-2)^2-1$

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal or "phase" shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\
\end{array}$

$\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{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
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now.. with that template above in mind, let's see yours

$\bf \begin{array}{lllllll}
f(x)=&3tan(&4x+&\pi )\\
f(x)=&3tan(&4x+&1\pi )&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

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