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LuckyWell [14K]
3 years ago
10

How to covert this y=(x-4)(x-12) into standard form

Mathematics
1 answer:
frez [133]3 years ago
6 0

Answer:

x^2-16x+48

Step-by-step explanation:

Use foil,

x^2-12x-4x+48

x^2-16x+48

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Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?
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bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.


\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1

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Answer: I believe its 50 because you would have to see 3 x what equals 150

Step-by-step explanation:

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