Given APWVS ~ HDTFY, what is HD?

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)

Step-by-step 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)