DETERMINE A QUADRATIC EQUATION GIVEN THE ROOTS AND A POINT

Problem 1 :

The roots of a quadratic equation are -2 and -6. The minimum point of the graph of its related function is at (-4, -2). Sketch the graph of the function.

Solution :

α = -2 and β = -6

α + β = -2 - 6 ==> -8

α β = -2 (-6) ==> 12

The quadratic function will be,

y = a(x² - (-8)x + 12)

y = a(x² + 8x + 12)

Since the quadratic function is having a minimum point (-4, -2), the will pass through this point.

-2 = a((-4)² + 8(-4) + 12)

-2 = a(16 - 32 + 12)

-2 = a(-4)

a = 1/2

By applying the value of a, we get

y = (1/2)(x² + 8x + 12)

So, the required quadratic function is

y = (1/2)(x² + 8x + 12)

Problem 2 :

The roots of a quadratic equation are -6 and 0. The minimum point of the graph of its related function is at (-3, 4). Sketch the graph of the function.

Solution :

α = -6 and β = 0

α + β = -6 + 0 ==> -6

α β = -6 (0) ==> 0

The quadratic function will be,

y = a(x² - (-6)x + 0)

y = a(x² + 6x)

Since the quadratic function is having a minimum point (-3, 4), the will pass through this point.

4 = a((-3)² + 6(-3))

4 = a(9 - 18)

4 = a(-9)

a = -4/9

By applying the value of a, we get

y = (-4/9)(x² + 6x)

So, the required quadratic function is

y = (-4/9)(x² + 6x)

Problem 3 :

Which of the following x equations represent the parabola?

a)   y = 2(x − 2)(x + 1)    b) y = 2(x + 0.5)² − 4.5   c) y = 2(x − 0.5)² − 4.5

d) y = 2(x + 2)(x − 1)

Solution :

The roots are -1 and 2. That is, x = -1 and x = 2

The factors are (x + 1) and (x - 2).

The quadratic function is f(x) = a(x + 1) (x - 2)

Applying the point (0.5, -4.5), we get

f(0.5) = a(0.5 + 1) (0.5 - 2)

-4.5 = a(1.5) (-1.5)

-4.5 = a(-2.25)

a = 4.5/2.25

a = 2

f(x) = 2(x + 1) (x - 2)

So, option a is correct.

Problem 4 :

 x-intercepts of 12 and −6; passes through (14, 4)

Solution :

Given that x-intercepts are 12 and -6. That is, x = 12 and x = -6

The factor form is (x - 12) (x + 6)

The quadratic function is f(x) = a(x + 1) (x - 2)

Applying the point (14, 4), we get

f(14) = a(14 - 12) (14 + 6)

4 = a(2)(20)

4 = 40a

a = 4/40

a = 1/10

Applying the value of a, we get

f(x) = (1/10)(x + 1) (x - 2)

Problem 5 :

 x-intercepts of 9 and 1; passes through (0, −18)

Solution :

Given that x-intercepts are 9 and 1. That is, x = 9 and x = 1

The factor form is (x - 9) (x - 1)

The quadratic function is f(x) = a (x - 9) (x - 1)

Applying the point (0, -18), we get

f(0) = a (0 - 9) (0 - 1)

-18 = a(-9)(-1)

-18 = 9a

a = -18/9

a = -2

Applying the value of a, we get

f(x) = -2 (x - 9) (x - 1)

Problem 6 :

x-intercepts of −16 and −2; passes through (−18, 72)

Solution :

Given that x-intercepts are -16 and -2. That is, x = -16 and x = -2

The factor form is (x + 16) (x + 2)

The quadratic function is f(x) = a (x + 16) (x + 2)

Applying the point (-18, 72), we get

f(-18) = a(-18 + 16) (-18 + 2)

72 = a(-2)(-16)

72 = 32a

a = 72/32

a = 2.25

Applying the value of a, we get

f(x) = 2.25 (x + 16) (x + 2)

Problem 7 :

Describe and correct the error in writing an equation of the parabola.

Solution :

By observing the x-intercepts, those are -1 and 2

Factored form is (x + 1)(x - 2)

y = a(x + 1)(x - 2)

The parabola passes through the point (3, 4), we get

4 = a(3 + 1)(3 - 2)

4 = a(4)(1)

4 = 4a

a = 4/4

a = 1

Applying the value of a, we get

y = 1(x + 1)(x - 2)