Quadratics assignment
Introduction
The aim of this investigation is to identify the nature of the roots of quadratics and cubic functions.
Part One
Case One
For Case One, the discriminant of the quadratic will always be equal to zero. This will result in the parabola cutting the axis once, or twice in the same place, creating a distinct root or two of the same root.
For PROOF 1, the equation y=a(x-b)2 is used.
PROOF 1 y = 3 (x – 2)2
= y = 3 (x2 – 4x + 4)
= y = 3x2 – 12x + 12
^ = b2 – 4ac
= (-12)2 – 4 x 3 x 12
= 144 – 144
= 0
The discriminant is equal to zero. The parabola touches the x-axis at (2, 0) and as a in the equation is a positive value, the parabola curves upwards.
For PROOF 2, the equation y=-a(x-b)2 is used.
PROOF 2 y = -3(x - 2)2
= y = -3 (x2- 4x + 4)
= y = -3x2 + 12x – 12
^ = b2 – 4ac
= 122 – 4 x (-3) x (-12)
= 144 – 144
= 0
The discriminant of the parabola is equal to zero. The parabola touches the x-axis at (2,0) and as a is a negative value, the parabola curves downwards.
For PROOF 3, the equation y=a(x-A)2 is used. y = a (x – A)2
= y = a (x2 – 2Ax + A2)
= y = ax2 – 2aAx + aA2
^ = b2 – 4ac
= (- 2aA)2 – 4 x (a) x (aA2)
= 4a2A2 - 4a2A2
= 0
The discriminant of the parabola is equal to zero. The parabola touches the x- axis at (A, 0) and whether the parabola curves up or down is dependent on the whether A is a positive or negative value.
Case Two
In Case Two, the discriminant of the quadratic will be equal to a positive number. Therefore the quadratic will cut the x – axis twice and this means that it will always have two real roots or zeros.
For PROOF 1, the equation y = a(x – b)(x – c) is used.
PROOF 1 y = 1 (x – 2)(x – 3)
= y = 1 (x2 – 5x + 6)
= y = 1x2 – 5x + 6)
^ = b2 – 4ac
= (-5)2 - 4 x 1 x 6
= 25 – 24
= 1
The discriminant of this equation is equal to 1. The parabola touches the x – axis at (2, 0) and (3, 0) and because a is a posotive, the paranola curves upwards. From this it can be
The aim of this investigation is to identify the nature of the roots of quadratics and cubic functions.
Part One
Case One
For Case One, the discriminant of the quadratic will always be equal to zero. This will result in the parabola cutting the axis once, or twice in the same place, creating a distinct root or two of the same root.
For PROOF 1, the equation y=a(x-b)2 is used.
PROOF 1 y = 3 (x – 2)2
= y = 3 (x2 – 4x + 4)
= y = 3x2 – 12x + 12
^ = b2 – 4ac
= (-12)2 – 4 x 3 x 12
= 144 – 144
= 0
The discriminant is equal to zero. The parabola touches the x-axis at (2, 0) and as a in the equation is a positive value, the parabola curves upwards.
For PROOF 2, the equation y=-a(x-b)2 is used.
PROOF 2 y = -3(x - 2)2
= y = -3 (x2- 4x + 4)
= y = -3x2 + 12x – 12
^ = b2 – 4ac
= 122 – 4 x (-3) x (-12)
= 144 – 144
= 0
The discriminant of the parabola is equal to zero. The parabola touches the x-axis at (2,0) and as a is a negative value, the parabola curves downwards.
For PROOF 3, the equation y=a(x-A)2 is used. y = a (x – A)2
= y = a (x2 – 2Ax + A2)
= y = ax2 – 2aAx + aA2
^ = b2 – 4ac
= (- 2aA)2 – 4 x (a) x (aA2)
= 4a2A2 - 4a2A2
= 0
The discriminant of the parabola is equal to zero. The parabola touches the x- axis at (A, 0) and whether the parabola curves up or down is dependent on the whether A is a positive or negative value.
Case Two
In Case Two, the discriminant of the quadratic will be equal to a positive number. Therefore the quadratic will cut the x – axis twice and this means that it will always have two real roots or zeros.
For PROOF 1, the equation y = a(x – b)(x – c) is used.
PROOF 1 y = 1 (x – 2)(x – 3)
= y = 1 (x2 – 5x + 6)
= y = 1x2 – 5x + 6)
^ = b2 – 4ac
= (-5)2 - 4 x 1 x 6
= 25 – 24
= 1
The discriminant of this equation is equal to 1. The parabola touches the x – axis at (2, 0) and (3, 0) and because a is a posotive, the paranola curves upwards. From this it can be