Given the following three points, find by hand the quadratic function they represent.(-1,-8), (0, -1),(1,2)(1 point)Of(x) =

7 answers
Question:

Given the following three points, find by hand the quadratic function they represent. (-1,-8), (0, -1),(1,2)
(1 point)
Of(x) = -51% + 87 - 1
O f(x) = -3.2? + 4.1 - 1
Of(t) = -202 + 5x - 1
Of(1) = -3.1? + 10.1 - 1​

Answers

The correct option is;

f(x) = -2·x² + 5·x - 1

Step-by-step explanation:

Given the points

(-1, -8), (0, -1), (1, 2), we have;

The general quadratic function;

f(x) = a·x² + b·x + c

From the given points, when x = -1, y = -8, which gives

-8 = a·(-1)² + b·(-1) + c = a - b + c

-8 =  a - b + c(1)

When x = 0, y = -1, which gives;

-1 = a·0² + b·0 + c = c

c = -1(2)

When x = 1, y = 2, which gives;

2 = a·1² + b·1 + c = a + b + c(3)

Adding equation (1) to (3), gives;

-8 + 2 = a - b + c + a + b + c

-6 = 2·a + 2·c

From equation (2), c = -1, therefore;

-6 = 2·a + 2×(-1)

-2·a  = 2×(-1)+6 = -2 + 6 = 4

-2·a = 4

a = 4/-2 = -2

a = -2

From equation (1), we have;

-8 =  a - b + c = -2 - b - 1 = -3 - b

-8 + 3 = -b

-5 = -b

b = 5

The equation is therefore;

f(x) = -2·x² + 5·x - 1

The correct option is f(x) = -2·x² + 5·x - 1.

(1) B

(2) D

Step-by-step explanation:

(1)

Let the quadratic function be:

[tex]y = ax^{2} + bx + c[/tex]

For the point, (0,-1),

[tex]y = ax^{2} + bx + c[/tex]

[tex]-1=(a\times0)+(b\times0}+c\\-1=c\\c=-1[/tex]

Then the equation is:

[tex]y = ax^{2} + bx -1[/tex]

For the point (-1, -8) ,

[tex]y = ax^{2} + bx -1[/tex]

[tex]-8=(a\times (-1)^{2})+(b\times -1)-1\\-8=a-b-1\\a-b=-7...(i)[/tex]

For the point (1, 2) ,

[tex]y = ax^{2} + bx -1[/tex]

[tex]2=(a\times (1)^{2})+(b\times 1)-1\\2=a+b-1\\a+b=3...(ii)[/tex]

Add the two equations and solve for a as follows:

[tex]a-b=-7\\a+b=3\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\2a = -4\\a = -2[/tex]

Substitute a = -2 in (i) and solve for b as follows:

[tex]a-b=-7\\-2-b=-7\\b=5[/tex]

Thus, the quadratic function is:

[tex]f(x)=-2x^{2}+5x-1[/tex]

The correct option is (b).

(2)

The ordered pairs are:

(5, 7), (7, 11), (9, 14), (11, 18)

Represent them in an XY table as follows:

X : 5 | 7 | 9 | 11

Y : 7 | 11 | 14 | 18

Compute the difference between the Y values as follows:

Diff = 11 - 7 = 4

Diff = 14 - 11 = 3

Diff = 18 - 14 = 4

Now compute the difference between the Diff values:

d = 3 - 4 = -1

d = 4 - 3 = 1

Since the differences between the differences of the y-values is not consistent, the ordered pairs do not represent a quadratic equation.

The correct option is D.

1) f(x) = 4·x² - 3·x + 6

2) f(x) = -2·x² + 5·x - 1

3) y = 2·(x - 3)² + 5

Step-by-step explanation:

1) The quadratic function that is represented by the points (0, 6), (2, 16), (3, 33) is found as follows

The general form of a quadratic function is f(x) = a·x² + b·x + c

Where, in (x, y), f(x) = y, and x = x

Therefore for the point (0, 6), we have;

6 = 0·x² + 0·x + c

c = 6

We have c = 6

For the point (2, 16), we have;

16 = a·2² + b·2 + 6

10 = 4·a + 2·b(1)

For the point (3, 33), we have;

33 = a·3² + b·3 + 6

27 = 9·a + 3·b(2)

Multiply equation (1) by 1.5 and subtract it from equation (2), we have;

1.5 × (10 = 4·a + 2·b)

15 = 6·a + 3·b

27 = 9·a + 3·b - (15 = 6·a + 3·b) gives;

27 - 15 = 9·a - 6·a+ 3·b - 3·b

12 = 3·a

a = 12/3 = 4

a = 4

From equation (1), we have;

10 = 4·a + 2·b = 4×4 + 2·b

10 - 4×4 = 2·b

10 - 16 = 2·b

-6 = 2·b

b = -3

The function, f(x) = 4·x² - 3·x + 6

2) Where the points are (-1, -8), (0, -1), (1, 2), we have;

For point (-1, -8), we have -8 = a·(-1)² - b·(-1) + c = a - b + c(1)

For point (0, 1), we have -1 = a×0² + b×0 + c = c(2)

For point (1, 2), we have 2 = a×1²+ b×1 + c = a + b + c(3)

Adding equation (1) to equation (3) gives

-8 + 2 = a - b + c +  a + b + c = 2·a + 2·c  where, c = -1, we have

-8 + 2 = -6 = 2·a + 2

2·a = -6 + 2 = - 4

a = -8/2 = -2

From equation (3), we have;

2 = a + b + c

b = 2 - a - c = 2 - (-2) - (-1) = 2 + 2 + 1 = 5

f(x) = -2·x² + 5·x - 1

3) The equation of a parabola that has vertex (3, 5) and passing through the point (1, 13) is given by the vertex equation of a parabola

The vertex equation of a parabola is y = a(x - h)² + k

Where;

(h, k) = Vertex (3, 5)

(x, y) = (1, 13)

We have

13 = a·(1 - 3)² + 5

13 = a·(-2)² + 5

13 - 5 = a·(-2)² = 4·a

4·a = 8

a = 8/4 = 2

The equation is y = 2·(x - 3)² + 5.

A B A C C are the correct answers~_~

Step-by-step explanation:

let the function be [tex]y=ax^2+bx+c[/tex]

put $x=0, \, y=6$ , to get $c=6$

put $x=2, \, y=16$ , $16=4a+2b+6\implies 2a+b=5$

put $x=3, \, y=33$ , $33=9a+3b+6\implies 3a+b=9$

subtract the two equation, to get $a=4$

now substitute $a$ in first equation, to get $b=5-2\cdot4=-3$

so, $f(x)=4x^2-3x+6$

The equation is;

f(x) = -2·x² + 5·x - 1

Step-by-step explanation:

The general form of a quadratic equation  or function f(x) is, f(x) = y = a·x² + b·x + c

Given that the points representing the quadratic function are;

(-1, -8), (0, -1), (1, 2) which  are of the form (x, y)

When x = -1, f(x) = y = -8

Plugging in the above values into the general form of a quadratic function, we have;

-8 = a·(-1)² + b·(-1) + c = a - b + c

-8  = a - b + c(1)

When x = 0, y = -1, we have;

-1 = a·(0)² + b·(0) + c = c

c = -1(2)

When x = 1, y = 2, which gives;

2 = a·(1)² + b·(1) + c = a + b + c

2 = a + b + c(3)

Adding equation (1) to equation (3), we have;

-8 + 2 = a - b + c + a + b + c

-8 + 2 = 2·a + 2·c

From equation (2) c = -1, we get;

-8 + 2 = -6 = 2·a + 2·c = 2·a + 2 × (-1)

-6 = 2·a - 2

-4 = 2·a

a = -2

From equation (3), we have

2 = a + b + c

Substituting the values of a, and c gives;

2 = -2 + b - 1

b = 2 + 2 + 1 = 5

b = 5

The equation is therefore;

f(x) = -2·x² + 5·x - 1.

A B A C C are the answers :)

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