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Marina86 [1]
3 years ago
10

The graph shows y as a function of x. Suppose a point is added to this graph. Which choice gives a point that preserves the func

tion, meaning it will keep the graph a function ?
A. (1,9)
B. (8,9)
C. (-3,3)
D. (-6,5)

Mathematics
2 answers:
sveticcg [70]3 years ago
8 0

Answer:

Option A. (1,9)

Step-by-step explanation:

we know that

A<em><u> function</u></em> is a relation from a set of inputs (independent variable) to a set of possible outputs (dependent variable) where <em><u>each input is related to exactly one output</u></em>.

so

<u><em>Verify each case</em></u>

case a) (1,9)

This point preserves the function

because, each input is related to exactly one output

For x=1, y=9

case b) (8,9)

This point not preserves the function

because, each input is not related to exactly one output

For x=8, y=9 and y=-3

case c) (-3,3)

This point not preserves the function

because, each input is not related to exactly one output

For x=-3, y=3 and y=4

case d) (-6,5)

This point not preserves the function

because, each input is not related to exactly one output

For x=-6, y=5 and y=-3

patriot [66]3 years ago
5 0

Answer:

Option A. (1,9)

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A N S W E R Q U I C K P L E A S E
chubhunter [2.5K]

Answer:

1. A

2. D

3. D

Step-by-step explanation:

The standard form of a parabola is

y=\frac{1}{4p}(x-h)^2+k            ..... (1)

Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.

1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .

k-p=-8                   ...(a)

(h,k+p)=(-2,-6)

k+p=-6            .... (b)

h=-2

On solving (a) and (b),  we get k=-7 and p=1.

Put h=-2, k=-7 and p=1 in equation (1).

y=\frac{1}{4(1)}(x-(-2))^2+(-7)

y=\frac{1}{4}(x+2)^2-7

Therefore option A is correct.

2 The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .

k-p=5                   ...(c)

(h,k+p)=(2,1)

k+p=1            .... (d)

h=2

On solving (c) and (d),  we get k=3 and p=-2.

Put h=2, k=3 and p=-2 in equation (1).

y=\frac{1}{4(-2)}(x-(2))^2+(3)

y=-\frac{1}{8}(x-2)^2+3

Therefore option D is correct.

3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .

k-p=-3                   ...(e)

(h,k+p)=(0,-2)

k+p=-2            .... (f)

h=0

On solving (e) and (f),  we get k=-2.5 and p=0.5.

Put h=0, k=-2.5 and p=0.5 in equation (1).

y=\frac{1}{4(0.5)}(x-(0))^2+(-2.5)

y=\frac{1}{2}(x)^2-2.5

y=\frac{1}{2}(x)^2-\frac{5}{2}

Therefore option D is correct.

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schepotkina [342]

Answer:

y=-4/7

Step-by-step explanation:

7y/2=-2

multiply both sides by 2

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divide both sides by 7

y=-4/7

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