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OLga [1]
1 year ago
13

How does the graph of g(x)=

Mathematics
1 answer:
kiruha [24]1 year ago
7 0

Answer:

Og(x) is shifted 4 units left and 6 units down from f(x).

Step-by-step explanation:

To understand how the parent function is transformed, you have to look at a few things.

Firstly, is there a negative sign in front? If there is, then the function is flipped around the y-axis

Second, on the part where the x is included (in this case it is x+4) you have to see if there is a negative sign in front of this. If this is the case, then the formula is flipped around the x-axis

<em>Third, If the part with the x is being added to, then the graph is being translated to the left that many units. If it is being subtracted from, then it is being translated to the right that many units (in this case it is </em><u><em>x+4</em></u><em>, so we move to the left 4 units) ((it is the opposite of what would be common sense, I know))</em>

<em>Lastly, if the whole thing is being added to, move up that many units. If it is subtracted from, move down that many units (in this case it is 1/x+4 </em><u><em>- 6)</em></u><em> (( this one does follow common sense))</em>

There are other factors, such as leading coefficients (on just the x part or the whole thing) and other stuff I'm sure I don't remember )

For more information: https://mathhints.com/parent-graphs-and-transformations/

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\bf \textit{area of this trapezoid}\\\\&#10;A=\cfrac{AB(BC+AD)}{2}\\\\&#10;-------------------------------\\\\&#10;\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -1}}\quad ,&{{ 5}})\quad &#10;%  (c,d&#10;B&({{ 3}}\quad ,&{{ 2}})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}&#10;\\\\\\&#10;AB=\sqrt{[3-(-1)]^2+[2-5]^2}\implies AB=\sqrt{(3+1)^2+(2-5)^2}&#10;\\\\\\&#10;AB=\sqrt{16+9}\implies AB=\sqrt{25}\implies \boxed{AB=5}

\bf -------------------------------\\\\&#10;\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;B&({{ 3}}\quad ,&{{ 2}})\quad &#10;%  (c,d&#10;C&({{ 0}}\quad ,&{{ -2}})&#10;\end{array}&#10;\\\\\\&#10;BC=\sqrt{(0-3)^2+(-2-2)^2}\implies BC=\sqrt{9+16}&#10;\\\\\\&#10;BC=\sqrt{25}\implies \boxed{BC=5}

\bf -------------------------------\\\\&#10;\textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;A&({{ -1}}\quad ,&{{ 5}})\quad &#10;%  (c,d&#10;D&({{ -13}}\quad ,&{{ -11}})&#10;\end{array}&#10;\\\\\\&#10;AD=\sqrt{[-13-(-1)]^2+[-11-5]^2}&#10;\\\\\\&#10;AD=\sqrt{(-13+1)^2+(-16)^2}\implies AD=\sqrt{144+256}&#10;\\\\\\&#10;AD=\sqrt{400}\implies \boxed{AD=\sqrt{20}}

so, the area for this trapezoid is then

\bf A=\cfrac{5(5+20)}{2}\implies A=\cfrac{125}{2}\implies A=62\frac{1}{2}

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