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mestny [16]
3 years ago
7

Function rule and y intercept

Mathematics
1 answer:
stiks02 [169]3 years ago
6 0

to get the equation for any straight line, all we need is two points from it, so hmmm say from this table let's use (2,0) and (6,-8)

\bf \begin{array}{|r|r|r|r|r|r|} \cline{1-6} days&x&2&4&6&8\\ \cline{1-6} dollars&y&0&-4&-8&-12\\ \cline{1-6} \end{array}~\hfill (\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-8}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{2}}}\implies \cfrac{-8}{4}\implies \cfrac{\stackrel{rise}{-2}}{\underset{run}{1}}\implies \stackrel{slope}{-2}

\bf \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}{0}=\stackrel{m}{-2}(x-\stackrel{x_1}{2}) \\\\\\ \stackrel{~\hfill y-intercept}{y=-2x\stackrel{\downarrow }{+4}}\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}

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algol [13]

Answer:

\boxed{\sf{4}}}

Step-by-step explanation:

Use a slope form.

<u>Slope:</u>

\sf{\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{rise}{run}  }

y₂ = (17)

y₁ = (1)

x₂ = (2)

x₁ = (-2)

Then, rewrite the problem.

Solve.

\sf{\dfrac{17-1}{2-(-2)}=\dfrac{16}{4}=\dfrac{16\div4}{4\div4}=\dfrac{4}{1}=\boxed{\sf{4}    }

Therefore, the slope is 4, which is our answer.

I hope this helps you! Let me know if my answer is wrong or not.

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