Question

If you were to write an equation of a parabola with x-intercepts 2 and −8 and pass through the point (−6 ,−4) , what would the a-value of the question be.

Answer 1

Answer:

$a=\frac{1}{4}$

Step-by-step explanation:

Quadratic equation (parabola) with x-intercepts x₁ and x₂:

$\boxed{y=a(x-x_1)(x-x_2)}$

Given:

x₁ = 2

x₂ = -8

(x, y) = (-6, -4)

$y=a(x-2)(x-(-8))$

$y=a(x-2)(x+8)$

Since y = a(x-2)(x+8) passes through point (-6, -4), we can substitute x and y with -6 and -4, then the equation will become:

$-4=a(-6-2)(-6+8)$

$-4=a(-8)(2)$

$a=-4\div(-16)$

  $=\frac{1}{4}$

Answer 2

Answer:

a =  $\frac{1}{4}$

Step-by-step explanation:

given a parabola with x- intercepts x = a and x = b, then the factors are

(x - a) and (x - b)

the equation of the parabola is then the product of the factors, that is

y = a(x - a)(x - b) ← a is a multiplier

given

x- intercepts x = 2 and x = - 8 , then the factors are

(x - 2) and (x - (- 8) ) , that is (x - 2) and (x + 8) , then

y = a(x - 2)(x + 8)

to find a, substitute (- 6, - 4 ) for x and y in the equation

- 4 = a(- 6 - 2)(- 6 + 8)

- 4 = a(- 8)(2) = - 16a ( divide both sides by - 16 )

$\frac{-4}{-16}$ = a ⇒ a = $\frac{1}{4}$