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timama [110]
3 years ago
14

If you are given the graph of h(x)=log6x. how could you graph m(x)=log6(x+3)

Mathematics
2 answers:
Dovator [93]3 years ago
8 0

We have been given the function h(x)=\log_6x

Now, we can see that in the transformed function  h(x)=\log_6(x+3), we have added 3 in the x.

Whenever we add/subtract some constant in the x, then the function gets translated to either left or right.

We know that if f(x) is a parent function and if we add a constant c then the function gets shift to c unit left.

Therefore, the given graph will get shifted 3 units left.

D is the correct option.

Translate each point of the graph of h(x) 3 units left.

Setler [38]3 years ago
4 0

Answer:it is D 3 unites left

Step-by-step explanation:

I took the test

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inysia [295]
If the diameter is at (9,4) and (-3,-2), then the center of the circle is the midpoint of that segment.

and since we know that the radius of a circle is half of the diameter, whatever long that diameter segment is, the radius is half that.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 9 &,& 4~) 
%  (c,d)
&&(~ -3 &,& -2~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
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\left( \cfrac{-3+9}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{center}{(3~,~1)}

\bf -------------------------------\\\\
~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 9 &,& 4~) 
%  (c,d)
&&(~ -3 &,& -2~)
\end{array}
\\\\\\
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2}
\\\\\\
d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}

\bf -------------------------------\\\\
\textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad 
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\\\\\\
(x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2}
\\\\\\
(x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45
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