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Sidana [21]
2 years ago
9

What is the domain and range of quadratic function with vertex.

Mathematics
1 answer:
Alexus [3.1K]2 years ago
4 0

Answer:

As with any function, the domain of a quadratic function f(x) is the set of x -values for which the function is defined, and the range is the set of all the output values (values of f ). Quadratic functions generally have the whole real line as their domain: any x is a legitimate input. The vertex will be the minimum (lowest point) of the graph if and the maximum (highest point) of the graph if

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Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo
noname [10]
Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
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%  (c,d)
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\end{array}\qquad
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now, a perpendicular line to RS, will have a negative reciprocal slope to it.  Well, what is the slope of RS anyway?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
%   (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
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and let's check the reciprocal negative of that,

\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies  +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

\bf \begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
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\\\\\\
% point-slope intercept
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\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}
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