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ch4aika [34]
2 years ago
6

Write the equation of the parabola in vertex form.

Mathematics
1 answer:
stellarik [79]2 years ago
8 0

well, looking at the picture of this vertically opening parabola, it has a vertex at 0,0 and it passes through 2,1 hmm ok

~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y = a(x-0)^2+0\qquad \stackrel{\textit{we also know that}}{x=2\qquad y = 1}\qquad \implies 1=a(2-0)^2+0 \\\\\\ 1=4a\implies \cfrac{1}{4}=a~\hspace{10em} \boxed{y=\cfrac{1}{4}x^2}

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4. Which of the following lines is perpendicular to y = -2X + 8?
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For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1.

That is to say:

m_ {1} * m_ {2} = - 1

We have the following equation:

y = -2x + 8

So:

m_ {1} = - 2

Thus:

m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {- 2}\\m = \frac {1}{2}

Thus, a line perpendicular to the given line must have slope m = \frac {1} {2}.

Option A:

x + 2y = 8\\2y = -x + 8\\y = - \frac {1} {2} x + 4

It is not perpendicular!

Option B:

x-2y = 6\\2y = x-6\\y = \frac {1} {2} x-3

If it is perpendicular!

Option C:

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It is not perpendicular!

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It is not perpendicular!

The correct option is option B

ANswer:

Option B

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