Question

Write a quadratic relation in the form y = ax2 + bx + c with roots 4 and - 3 and

passing through the point (3, -12).

Answer 1

y = 2x² - 2x - 24

Explanation:

Given a root x = a then the factor is (x - a )

Given roots are x = 4 and x = - 3 , the corresponding factors are

(x - 4) and (x - (- 3)) , that is (x - 4) and (x + 3)

The quadratic is then the product of the factors

y = a(x - 4)(x + 3) ← a is a multiplier

To find a substitute (3, - 12) into the equation

- 12 = a(- 1)(6) = - 6a ( divide both sides by - 6 )

2 = a

y = 2(x - 4)(x + 3) ← expand factors using FOIL

= 2(x² - x - 12) ← distribute

y = 2x² - 2x - 24

Answer 2

2x²-2x-24

Explanation:

Something has roots of 4 and -3 we can write

(x-4)(x+3)

We then attach a constant, a that will ensure that it passes through the correct point

a(x-4)(x+3)

now plug in the numbers and solve for a

a(3-4)(3+3)= -12

a(-1)(6)= -12

-6a= -12

a=2

So we have

2(x-4)(x+3)

and now it's just a matter of mulitplying/simplifying things

(x-4)(x+3)= x²-x-12

2(x²-x-12)= 2x²-2x-24