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gladu [14]
3 years ago
7

XYZ has coordinates X(2, 3), Y(1,4), and Z(8,9). A translation maps X to X'(4,7). What are the coordinates for Y' and Z' for thi

s translation?​
Mathematics
1 answer:
Rashid [163]3 years ago
7 0

Answer:

The coordinates are Y'(x,y) = (3, 8) and Z'(x,y) = (10, 13).

Step-by-step explanation:

First, we have to derive an expression for translation under the assumption that each point of XYZ experiments the same translation. Vectorially speaking, translation from X to X' is defined by:

X'(x,y) = X(x,y) + T(x,y) (1)

Where T(x,y) is the vector translation.

If we know that X(x,y) = (2,3) and X'(x,y) = (4,7), then the vector translation is:

T(x,y) = X'(x,y)-X(x,y)

T(x,y) = (4,7) - (2,3)

T(x,y) = (2, 4)

Then, we determine the coordinates for Y' and Z':

Y'(x,y) = Y(x,y) + T(x,y)

Y'(x,y) = (1,4) + (2,4)

Y'(x,y) = (3, 8)

Z'(x,y) = Z(x,y) + T(x,y)

Z'(x,y) =(8,9) + (2,4)

Z'(x,y) = (10, 13)

The coordinates are Y'(x,y) = (3, 8) and Z'(x,y) = (10, 13).

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Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20
vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of y=f(x) is a vertical has an asymptote at x=a if at least one of the following statements is true:

1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty

The function is:

f(x)=\frac{x^2+7x+10}{x^2+9x+20}

First of all, let't factor out:

f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq  5

From here:

\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty

\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty

Accordingly:

x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

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