How to solve number 13

Mathematics

1 answer:

noname [10]3 years ago

8 0

$\bf \cfrac{3x-1}{x+2}-\cfrac{x-2}{x-1}\impliedby \stackrel{our~LCD~is}{(x+2)(x-1)} \\\\\\ \cfrac{(3x-1)(x-1)-\stackrel{\textit{difference of squares}}{(x-2)(x+2)}}{(x+2)(x-1)} \\\\\\ \cfrac{(3x^2-4x+1)-(x^2-2^2)}{(x+2)(x-1)}\implies \cfrac{(3x^2-4x+1)-(x^2-4)}{(x+2)(x-1)} \\\\\\ \cfrac{3x^2-4x+1-x^2+4}{x^2+x-2}\implies \cfrac{2x^2-4x+5}{x^2+x-2}$

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Find the vertical asymptotes. 2x2 + 7x + 6 y = 3x2 + 10x - 8 * = [ [?], x=

Llana [10]

Answer:

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Step-by-step explanation:

1. rewrite the equation in standard form: $4\cdot \frac{3}{2}\left(y-\left(-\frac{41}{24}\right)\right)=\left(x-\left(-\frac{3}{2}\right)\right)^2$

2. find (h,k), the vertex. the vertex is $\left(h,\:k\right)=\left(-\frac{3}{2},\:-\frac{41}{24}\right)$

3. find the 'focal length' of the parabola - the focal length is the distance between the vertex and the focus. from the vertex we can see that the focal length, p, = 3/2

4. Parabola is symmetric around the y-axis and so the asymptote is a line parallel to the x-axis, a distance p from the $\left(-\frac{3}{2},\:-\frac{41}{24}\right)$ y coordinate which is at $-\frac{41}{24}\right)$ . Set up the equation: