# FIND QUADRATIC EQUATION FROM X INTERCEPTS AND A  POINT

If x1 and xare x-intercepts, then the given x-intercepts can be written as x = x1and x = x2.

The quadratic equation can be derived using

y = a (x - x1) (x - x2)

By applying the given point as (x, y) in the equation above, we can derive the value of a.

Write a quadratic function in intercept form whose graph has the given x- intercepts and passes through the given point.

Problem 1 :

x - Intercepts : 2, 3

Point : (4, 2)

Solution :

y = a(x - x1) (x - x2)

Here x1   = 2, x2 = 3

y = a(x - 2) (x - 3) --- > (1)

Substitute x = 4 and y = 2

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

2 = a(2)(1)

2a = 2

a = 1

By applying a = 1 in (1)

y = 1(x - 2) (x - 3)

y = x² - 3x - 2x + 6

y = x² - 5x + 6

Problem 2 :

x - Intercepts : -4, 1

Point : (-3, -4)

Solution :

y = a(x - x1) (x - x2)

Here x1 = -4, x2 = 1

y = a(x + 4) (x - 1) --- > (1)

Substitute x = -3 and y = -4

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

-4 = a(1)(-4)

-4 = -4a

a = 1

By applying a = 1 in (1)

y = 1(x + 4) (x - 1)

y = x² - x + 4x - 4

y = x² + 3x - 4

Problem 3 :

x - Intercepts : -5, 5

Point : (6, 11)

Solution :

y = a(x - x1) (x - x2)

Here x1 = -5, x2 = 5

y = a(x + 5) (x - 5) --- > (1)

Substitute x = 6 and y = 11

11 = a(6 + 5) (6 - 5)

-11 = a(11)(1)

-11 = 11a

a = -1

By applying a = -1 in (1)

y = -1(x + 5) (x - 5)

y = -(x² - 5x + 5x - 25)

y = -x² + 25

y = x² - 25

Problem 4 :

x - Intercepts : -7, -2

Point : (-5, -6)

Solution :

y = a(x - x1) (x - x2)

Here x1 = -7, x2 = -2

y = a(x + 7) (x + 2) --- > (1)

Substitute x = -5 and y = -6

-6 = a(-5 + 7) (-5 + 2)

-6 = a(2)(-3)

-6 = -6a

a = 1

By applying a = 1 in (1)

y = 1(x + 7) (x + 2)

y = x² + 2x + 7x + 14

y = x² + 9x + 14

Problem 5 :

x - Intercepts : 0, 4

Point : (-1, 20)

Solution :

y = a(x - x1) (x - x2)

Here x1 = 0, x2 = 4

y = a(x - 0) (x - 4) --- > (1)

Substitute x = -1 and y = 20

20 = a(-1 - 0) (-1 - 4)

20 = a(-1)(-5)

20 = 5a

a = 4

By applying a = 4 in (1)

y = 4(x - 0) (x - 4)

y = 4(x)(x - 4)

y = 4(x² - 4x)

y = 4x² - 16x

Problem 6 :

x - Intercepts : -3, -2

Point : (-4, -6)

Solution :

y = a(x - x1) (x - x2)

Here x1 = -3, x2 = -2

y = a(x + 3) (x + 2) --- > (1)

Substitute x = -4 and y = -6

-6 = a(-4 + 3) (-4 + 2)

-6 = a(-1)(-2)

-6 = 2a

a = -3

By applying a = -3 in (1)

y = -3(x + 3) (x + 2)

y = -3(x² + 2x + 3x + 6)

y = -3(x² + 5x + 6)

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