A parabola passes through the three points......?

The equation of the parabola is:

ax² + bx + c = y

Plug in the three points and get three equations.

a - b + c = 1

a + b + c = -2

16a + 4b + c = 1

Subtract the first equation from the second.

2b = -3

b = -3/2

Plug the value for b back in the second and third equations.

a + c = -1/2

16a + c = 7

Subtract the first equation from the second.

15a = 15/2

a = 1/2

Solve for c.

1/2 + c = -1/2

c = -1

The equation of the parabola is:

y = (1/2)x² - (3/2)x - 1

Solve for the vertex of the parabola.

y + 1 = (1/2)(x² - 3x)

y + 1 + (1/2)(9/4) = (1/2)(x² - 3x + 9/4)

y + 17/8 = (1/2)(x - 3/2)²

The vertex is (h,k) = (3/2, -17/8).

Solve for the zeros of the parabola.

(1/2)x² - (3/2)x - 1 = 0

x² - 3x - 2 = 0

x = {3 ± √[(-3)² + 4*1*(-2)]} / 2 = {3 ± √[9 + 8]} / 2 = (3 ± √17) / 2

The zeros are:

x = (3 ± √17) / 2

y = ax^2 + bx + c

Substitute every point into the equation:

1 = a - b + c...........(1)

-2 = a + b + c.........(2)

1 = 16a + 4b + c.....(3)

Use equation 2 minus equation 1:

2b = -2-1

2b = -3

b = -3/2

Use equation 3 minus equation 1:

15a + 5b = 0

3a + b = 0

3a + (-3/2) = 0

3a = 3/2

a = 1/2

Substitute a = 1/2 and b = -3/2 into equation 1

1 = 1/2 - (-3/2) + c

c = -1

Therefore, y = 1/2 x^2 - 3/2 x - 1

When y = 0,

1/2 x^2 - 3/2 x - 1 = 0

x^2 - 3x - 2 = 0

x = (3 + Sqrt 17) / 2 or (3 - Sqrt 17) / 2

The answer of x is same as x-intercepts.

To find vertex, we need to do complement squares

1/2 x^2 - 3/2 x - 1

= 1/2 (x^2 - 3 x) - 1

= 1/2 [ (x - 3/2)^2 - 9/4 ] - 1

= 1/2 (x - 3/2)^2 - (1/2)(9/4) - 1

= 1/2 (x - 3/2)^2 - 9/8 - 1

= 1/2 (x - 3/2)^2 - 17/8

The vertex is (3/2, -17/8)

It is the minimum point

plug x and y in y=ax^2+bx+c

You get

a-b+c=1

a+b+c = -2

16a+4b+c=1

subtract the second from first

-2b = 3 then b = -3/2

substitute in the third

16a-6+c=1

16a+c = 7

a+c = -1/2

15a = 15/2

a = 1/2

b = -3/2

c = -1

f(x) = (1/2)x^2-(3/2)x-1