Given the following linear function sketch the graph of the function and find the domain and range.

Mathematics

1 answer:

barxatty [35]3 years ago

3 0

As with any odd-degree polynomial, ...
the domain is "all real numbers"
the range is "all real numbers"

Here is a graph.

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What is the equation for a circle with a center at (-2,-4) that passes through the point (3,8)?

Margarita [4]

Check the picture below.

so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -2 &,& -4~) 
% (c,d)
&&(~ 3 &,& 8~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{3-2}{2}~~,~~\cfrac{8-4}{2} \right)\implies \left( \frac{1}{2}~,~2 \right)\impliedby center$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -2 &,& -4~) 
% (c,d)
&&(~ 3 &,& 8~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2}
\\\\\\
d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad \stackrel{radius}{\frac{13}{2}}$

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{\frac{1}{2}}{ h},\stackrel{2}{ k})\qquad \qquad 
radius=\stackrel{\frac{13}{2}}{ r}
\\\\\\
\left( x-\frac{1}{2} \right)^2+(y-2)^2=\left( \frac{13}{2} \right)^2\implies \left( x-\frac{1}{2} \right)^2+(y-2)^2=\frac{169}{4}$