How do you sketch this on parabola

Mathematics

1 answer:

hammer [34]3 years ago

8 0

Answer:

Sketching Parabolas

Find the vertex.. ...

Find the y -intercept, (0,f(0)) ( 0 , f ( 0 ) ) .

Solve f(x)=0 f ( x ) = 0 to find the x coordinates of the x -intercepts if they exist. ...

Make sure that you've got at least one point to either side of the vertex. ...

Sketch the graph.

Step-by-step explanation:

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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{3}}x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill$

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-1}{3}} ~\hfill \stackrel{reciprocal}{\cfrac{3}{-1}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{3}{-1}\implies 3}}$

so we're really looking for the equation of a line whose slope is 3 and passes through (1 , 10)

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{10})\qquad \qquad \stackrel{slope}{m}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{3}(x-\stackrel{x_1}{1}) \\\\\\ y-10=3x-3\implies y=3x+7$