Question

2. Write a quadratic equation with roots (-2,0) and (4,0) that has a minimum value of -36

Answer 1

Answer:

f(x) = 4*(x+2)*(x-4) factorized formula

f(x) = 4x^2 - 8x - 32 polynomical formula

Step-by-step explanation:

Lets assume the factorized formula of a quadratic function:

f(x) = a*(x-x1)^2 * (x-x2)^2

where "a" is a coefficient and x1, x2 are the roots of the function. Then replacing the given roots:

f(x) = a*(x+2)*(x-4)

because its told us that -36 is the minimum value of the function we can say that its concave, then this value is actually the component in the Y axis of the vertex. To find the component in the X axis of the vertex we have to make the adding between the roots and divide this value by 2, this last is because a quadratic function is a symetrical function. Lets call the component at the X axis of the vertex as Xv, then:

Xv=(x1+x2)/2

Xv=(-2+4)/2

Xv=1

Therefore now we have a point of the function and its P=(1,-36) this point is the vertex of the function too.

Now the last thing to do is to find the value of the coefficient "a". We can find it by replacing the point of the vertex obtained before.

-36 = a*(1+2)*(1-4)

-36 = a*(3)*(-3)

-36 = a*(-9)

4 = a

Finally the equation is:

f(x) = 4*(x+2)*(x-4)

if we expand this function we find the polynomical form of this function

f(x) = 4x^2 - 8x - 32