Answer:
Part A
The equation of the parabola in standard form is f(x) = 0.5·x² + x - 7.5
Part B
The equation of the parabola in factored form is f(x) = 0.5·(x + 5)·(x - 3)
Explanation:
The coordinates of the vertex of the parabola, (h, k) = (-1, -8)
The zeros of the parabola = -5, 3
Part A
The quadratic equation representing the parabola can be written as follows;
y = a·(x - h)² + k
The standard form of the equation is f(x) = a·x² + b·x + c
k = c - b²/(4·a)
h = -b/(2·a)
c = k + a·h² = -8 + a
From the zeros of the parabola, we have;
f(-5) = a × (-5)² + b×(-5) + c = 25·a - 5·b + c = 0
f(3) = a × (3)² + b×(3) + c = 9·a + 3·b + c = 0
Therefore, we have;
25·a - 5·b + c = 0...(1)
9·a + 3·b + c = 0...(2)
From which we have;
25·a - 5·b + c = 25·a - 5·b - 8 + a = 26·a - 5·b - 8 = 0
9·a + 3·b + c = 9·a + 3·b - 8 + a = 10·a + 3·b - 8 = 0
We then have;
26·a - 5·b = 8...(3)
10·a + 3·b = 8...(4)
Making 'b', the subject of both equations (3) and (4), equating the values, we find the value of 'a' as follows;
From equation (3), we have;
26·a - 5·b = 8
∴ b = (26·a - 8)/5
From equation (4), we have;
10·a + 3·b = 8
∴ b = (8 - 10·a)/3
From b = b, by transitive property, we have;
(26·a - 8)/5 = (8 - 10·a)/3
3 × (26·a - 8) = 5 × (8 - 10·a)
78·a - 24 = 40 - 50·a
∴ 128·a = 64
a = 64/128 = 1/2 = 0.5
a = 0.5
∴ b = (26·a - 8)/5 = (26 × 0.5 - 8)/5 = (13 - 8)/5 = 1
b = 1
c = -8 + a
∴ c = -8 + 0.5 = -7.5
c = -7.5
The equation of the parabola in standard form is therefore presented as follows;
f(x) = 0.5·x² + x - 7.5
Part B
The equation of the parabola in factored form is given as follows;
With the aid of a graphing calculator, we have;
f(x) = 0.5·x² + x - 7.5 = 0.5×(x² + 2·x - 15) = 0.5·(x + 5)·(x - 3)
The equation of the parabola in factored form is f(x) = 0.5·(x + 5)·(x - 3)