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strojnjashka [21]
3 years ago
9

Find an equation in standard form of a parabola passing through the points below

Mathematics
1 answer:
EastWind [94]3 years ago
8 0

Answer:

Basic parabola:  y = ax2 + bx + c

We have 3 points we can plug in for (x, y) to create 3 simultaneous equations

(-2, 24):  24 = 4a - 2b + c   {equation 1}

(3, -1):   -1 = 9a + 3b + c    {equation 2}

(-1, 15): 15 = a - b + c        {equation 3}

Solve this system to find the values of a, b, c

Let's first eliminate variable c:

4a - 2b + c = 24   {equation 1}

 a -  b + c = 15    {equation 3}

---------------------  subtract

3a - b = 9

9a + 3b + c = -1    {equation 2}

 a -  b + c = 15    {equation 3}

--------------------  subtract

8a + 4b = -16

We now have two equations with 2 unknowns we can use to find a, b

3a - b = 9         {equation 4}

8a + 4b = -16   {equation 5}

Multiply equation 4 through by 4 and add equations

12a - 4b = 36

8a + 4b = -6

-----------------   add

20a = 30

a = 30/20

a = 3/2

8a + 4b = -6

8(3/2) + 4b = -6

12 + 4b = -6

4b = -6- 12

4b = -18

b = -18/4

b = -9/2

Plug these 2 values into one of the original equations and solve for c

15 = a - b + c        {equation 3}

15 = 3/2 + 9/2 + c

15 = 12/2 + c

15 = 6 + c

c = 15-6

c = 9

y = (3/2)x2 - (9/2)x+ 9

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Select the point that is a solution to the system of inequalities
Zarrin [17]
<h2>Hello!</h2>

The answer is:

The point that is a solution to the system of inequalities is C(4,2).

<h2>Why?</h2>

To find the point that is a solution to the system of inequalities, we need to evaluate it into the given inequalities. If the point is a solution to the system of inequalities, both inequalities will be satisfied.

We are given the inequalities:

y\leq 2x-2

y\leq x^{2} -3x

So, substituting the given points into the given inequalities, we have:

- A. (2,1)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\1\leq 2*(2)-2\\\\1\leq 4-2\\\\1\leq 2

Substituting into the second inequality, we have:

y\leq x^{2} -3x

1\leq (2)^{2} -3(2)

1\leq (2)^{2} -3(2)

1\leq 4 -6

1\leq -2

Therefore, since 1 is not less or equal to -2, the point A(2,1) is not a solution to the system of inequalities.

- B. (-2,-1)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\-1\leq (-2)*(2)-2\\\\-1\leq -4-2\\\\-1\leq -6

Substituting into the second inequality, we have:

y\leq x^{2} -3x

-1\leq (-2)^{2} -3(-2)

-1\leq 4 +6

-1\leq 10

Therefore, since -1 is not less or equal to -6, the point B(-2,-1) is not a solution to the system of inequalities.

- C. (4,2)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\2\leq (4)*(2)-2\\\\2\leq 8-2\\\\2\leq 6

Substituting into the second inequality, we have:

y\leq x^{2} -3x

2\leq (4)^{2} -3(4)

2\leq 16 -12

2\leq 4

Therefore, since 2 is less than 6, and 2 is less than 4, the point B(4,2) is a solution to the system of inequalities.

- D(1,3)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\3\leq (1)*(2)-2\\\\3\leq 2-2\\\\3\leq 0

Substituting into the second inequality, we have:

y\leq x^{2} -3x

3\leq (1)^{2} -3(1)

3\leq 1 -3

3\leq -2

Therefore, since 3 is not less than 0, and 3 is not less than -2, the point D(1.3) is not a solution to the system of inequalities.

Hence, the point that is a solution to the system of inequalities is C(4,2).

Have a nice day!

6 0
3 years ago
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