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2 years ago

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How is the graph of y=(x-1)2-3 transformed to produce the graph of y-3(x+4)??

1 answer:

Salsk061 [2.6K]2 years ago

4 0

Step-by-step explanation:

How is the graph of y=(x-1)2-3 transformed to produce the graph of y-3(x+4)??

The graph is translated left 5 units, compressed vertically by a factor of 2, and translated up 3 units.

The graph is stretched vertically by a factor of Ź, translated left 5 units, and translated up 3 units.

O The graph is translated left 5 units, compressed horizontally by a factor of 3, and translated down 3 units.

O The graph is stretched horizontally by a factor of Ž, translated left 5 units, and translated down 3 units.

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1 year ago

A quadrilateral has vertices at A(-5,5), B(1,8), C(4,2), and D(-2,-2). Use slope to determine if the quadrilateral is a rectangl

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Answer:

The quadrilateral is not a rectangle

Step-by-step explanation:

we know that

If a quadrilateral ABCD is a rectangle

then

Opposite sides are congruent and parallel and adjacent sides are perpendicular

Remember that

If two lines are parallel, then their slopes are the same

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

A(-5,5), B(1,8), C(4,2), and D(-2,-2)

Plot the figure to better understand the problem

see the attached figure

Find the slope of the four sides and then compare

step 1

Find slope AB

A(-5,5), B(1,8)

substitute in the formula

$m=\frac{8-5}{1+5}$

$m=\frac{3}{6}$

$m_A_B=\frac{1}{2}$

step 2

Find slope BC

B(1,8), C(4,2)

substitute in the formula

$m=\frac{2-8}{4-1}$

$m=\frac{-6}{3}$

$m_B_C=-2$

step 3

Find slope CD

C(4,2), and D(-2,-2)

substitute in the formula

$m=\frac{-2-2}{-2-4}$

$m=\frac{-4}{-6}$

$m_C_D=\frac{2}{3}$

step 4

Find slope AD

A(-5,5), D(-2,-2)

substitute in the formula

$m=\frac{-2-5}{-2+5}$

$m=\frac{-7}{3}$

$m_A_D=-\frac{7}{3}$

step 5

Verify if the opposites are parallel

Remember that

If two lines are parallel, then their slopes are the same

The opposite sides are

AB and CD

BC and AD

we have

$m_A_B=\frac{1}{2}$

$m_C_D=\frac{2}{3}$

so

$m_A_B \neq m_C_D$

It is not necessary to continue verifying, because two of the opposite sides are not parallel

therefore

The quadrilateral is not a rectangle

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2 years ago