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bearhunter [10]
1 year ago
5

Write the equation of the line that passes through the points (-6,-1) and (-4,2). Put your answer in fully simplified point-slop

e form, unless it is a vertical or horizontal line.
Mathematics
1 answer:
Archy [21]1 year ago
4 0

(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-6)}}} \implies \cfrac{2 +1}{-4 +6} \implies \cfrac{ 3 }{ 2 }

\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}{(-1)}=\stackrel{m}{\cfrac{3}{2}}(x-\stackrel{x_1}{(-6)}) \implies {\large \begin{array}{llll} y +1= \cfrac{3}{2} (x +6) \end{array}}

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The polynomial $f(x)$ has degree 3. If $f(-1) = 15$, $f(0)= 0$, $f(1) = -5$, and $f(2) = 12$, then what are the $x$-intercepts o
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Your calculator's cubic regression function can tell you the equation is

... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)

The x-intercepts are -4, 0, +1.5.

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Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.

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... a + b + c = -5 . . . . for x=1

... 8a +4b +2c = 12 . . for x=2

adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to

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... 4a +c = -4

Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).

Then your equation is

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Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.

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as the product of 90 degrees and 20 degrees

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