I'm doing a math sheet as an assignment due tomorrow and I'm stuck

Mathematics

1 answer:

Dafna1 [17]2 years ago

7 0

Hi, so if x=0 and if y=x+5 then you would simply add 5 to x to get y, so in this case y=5. Then if x=1, then y would be 1+5, y=6. With the second part, it's the same thing. If x=1 and y=2x+3, all you have to do is multiply 1 by 2 wich = 2, then you would add 3 to that, wich is 5. And if x=2 and y=2x+3, then you would multiply 2 by 2, wich is 4, then add 3 to that wich would be 7.I hope this helps, respond to this answer if you have any comments or questions ;D

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$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
~~~~y={{ A}}({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)={{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
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f(x)={{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
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\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if }{{ B}}\textit{ is negative}\\
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$\bf \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
~~~~~~if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{ D}}\\
~~~~~~if\ {{ D}}\textit{ is negative, downwards}\\\\
~~~~~~if\ {{ D}}\textit{ is positive, upwards}\\\\
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with that template in mind,

$\bf f(x)=x^2\qquad \quad f(x+2)=(x+2)^2\implies f(x+2)=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+2})^2\stackrel{D}{+0}$

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