Re-write the quadratic function below in Standard Form y= 8(x + 1)2 – 3

Mathematics

1 answer:

lubasha [3.4K]4 years ago

3 0

you will simplify the 1st term in the RHS so it will be

$y = 16x + 16 - 3$

you will simplify the like terms which will be

$y = 16x + 13$

This is the final answer because it is express in

y=mx+c

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If r= 3x-1 and s = 2x-1 how do i get this answer r/s(6)

e-lub [12.9K]

The answer is:

The answer is:

Step-by-step explanation:

5 0

3 years ago

What is the period and midline?

OverLord2011 [107]

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

\end{array}\qquad$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\
\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}\\
\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\


\end{array}$

$\bf \begin{array}{llll}
\bullet \textit{function period}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}
$

so if you notice yours $\bf \begin{array}{llll}
3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\
&\ \uparrow&\uparrow \\
&B&D 
\end{array}$

now.. normally the function $\bf 3.2cos&\left( \frac{5}{3}\theta \right)$
has a D value of 0, or no vertical shift, and the amplitude and the period simply make the wave taller and thinner, but the midline is still the x-axis

now, with D = 6.1, that moves the midline up vertically that much

now.. the period, well, B = 5/3, normal period of cosine is $2\pi$
so, the new period will be $\bf \cfrac{2\pi }{B}\implies \cfrac{2\pi }{\frac{5}{3}}\implies \cfrac{6\pi }{5}$

notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)