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Rudiy27
3 years ago
5

Also do the line pass through the origin? Explain.

Mathematics
1 answer:
monitta3 years ago
8 0

(\stackrel{x_1}{4}~,~\stackrel{y_1}{-1})\qquad \qquad m = -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}{(-1)}=\stackrel{m}{-3}(x-\stackrel{x_1}{4})\implies y+1=-3(x-4) \\\\\\ y+1=-3x+12\implies \stackrel{\textit{slope-intercept form}}{y=-3x+11}

does it pass through the origin?  well, we could draw it, or we can just check that, the origin is at 0,0, or namely when x = 0, y = 0, let's see if that's true, let's make x = 0, let's see what we get for "y".

y = -3(0) + 11 => y = 11

woops, no dice, y ≠ 0, so nope, doesn't pass through the origin.

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(-1, 1)

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um i kinda feel like i got it wrong buuuuuuut yea

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First, we should try to change the last number in a way so we can factor the entire thing

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tooo make it look like y=(x + 1 )²

the only difference between the original y=x² + 2x + 2

and the factor-able y=x² + 2x + 1 is simply changing the

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ummmmm yeeaaa <3

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