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2 years ago

What is the range of the graph below? I think it's B am i wrong?

Mathematics

2 answers:

nignag [31]2 years ago

8 0

I would believe that it is B as well. I am not the best with graphs. But, I did literally just cover this subject in math. Let me explain why I believe it is B.

Range, graph wise and functions wise, is the possible y values. The point goes down to -4. The points go up towards. Meaning that possible answers would have to be greater(or equal to) -4.

Does that make sense?

eduard2 years ago

6 0

Answer: B) {y | y ≥ -4}

Step-by-step explanation:

Range is: all the values that y can be. The lowest y-value of the given graph is -4 and the the highest y-value is positive infinity.

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
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\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\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{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template in mind, let's take a peek at this function

$\bf \begin{array}{lllcclll}
y=&2(&1x&-2)^2&-4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
A\cdot B=2\impliedby \textit{shrunk by a factor of 2, of half-size}\\\\
\cfrac{C}{B}= \cfrac{-2}{1}\implies -2\impliedby \textit{horizontal right shift of 2 units}\\\\
D=-4\impliedby \textit{vertical down shift of 4 units}$

so, the graph of y=2(x-2)²-4, is really the same graph of y=x², BUT, narrower, and moved about horizontally and vertically

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