Question

Writing Quadratic Equations

Answer 1

Explanation:

Given:

• vertex (p, q) = (-4, -4)
• roots (x₁, 0) & (x₂, 0) = (-2, 0) & (-6, 0)

`\boxed{vertex\ form:f(x)=a(x-p)^2+q}`

f(x) = a(x-(-4))² + (-4)

f(x) = a(x+4)² - 4

[in order to find value of a, substitute x & f(x) with any coordinate from the parabola, e.g.(-2, 0)]

0 = a(-2+4)² - 4

0 = 4a - 4

a = 1

Therefore:

vertex form : f(x) = (x+4)² - 4

To find the standard form, just expand the vertex form:

f(x) = (x+4)² - 4

f(x) = (x²+8x+16) - 4

f(x) = x² + 8x + 12

To find the intercept form, just factorize the standard form

f(x) = x² + 8x + 12

f(x) = (x+2)(x+6)