If x_{1} and x_{2 }are x-intercepts, then the given x-intercepts can be written as x = x_{1}and x = x_{2}.
The quadratic equation can be derived using
y = a (x - x_{1}) (x - x_{2})
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 - x_{1}) (x - x_{2})
Here x_{1 } = 2, x_{2} = 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 - x_{1}) (x - x_{2})
Here x_{1 }= -4, x_{2} = 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 - x_{1}) (x - x_{2})
Here x_{1 }= -5, x_{2} = 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 - x_{1}) (x - x_{2})
Here x_{1 }= -7, x_{2} = -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 - x_{1}) (x - x_{2})
Here x_{1 }= 0, x_{2} = 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 - x_{1}) (x - x_{2})
Here x_{1 }= -3, x_{2} = -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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