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3 years ago

7

For problems 1 – 6, graph each function. Label the vertex and axis of symmetry.

1 answer:

adelina 88 [10]3 years ago

4 0

1. $y=x^{2} -2x-1$

a= 1 , b= -2 , c= -1

$\frac{-b}{2a}$ $\frac{- (-2)}{2 times 1}$

x=1

now use the x value to find y

$y=1^{2} -2 times 1 - 1$

y = -2

vertex (1,-2)

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The coordinates of midpoint M and endpoint Q of a segment are M(–7, 8) and Q(–19, 30). Find the coordinates of the other endpoin

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Answer:

Step-by-step explanation:

Use the midpoint formula to solve:

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We know the midpoint and one set of coordinates, so filling those in:

$(-7,8)=(\frac{x+(-19)}{2},\frac{y+30}{2})$

Since the -7 is the x coordinate of the midpoint, we set the x part of the equation equal to -7 and solve for x, and then do the same for y. The 2 equations are:

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2 years ago

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3 years ago

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iogann1982 [59]

Answer:

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Reptile [31]

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Here, we have

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First, notice we have x in both the numerator and denomiator so we have a removable discounties at x.

Since, we don't want x to be 0,

We have a removable discontinuity at x=0

Now, we have

$\frac{2x - 3}{x + 2}$

We don't want the denomiator be zero because we can't divide by zero.

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All Real Numbers except-2 and 0.

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