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A quadratic relation has zeros 2 and 6, and it has an optimal value of 3. Determine the equation of the relation in factored form.

User JosephA
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1 Answer

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Answer:

y = (-3/4)*(x - 2)*(x-6)

Explanation:

Ok, we can write a quadratic equation as:

Y = a*(x - b)*(x - c)

where a is a scalar, b and c are the roots.

We know that b = 2 and c = 6, so we have:

y = a*(x - 2)*(x - 6)

now, we can expand this and get:

y = a*(x^2 -8x + 12)

The optimal value of this quadratic equatin is when:

x = 8/2 = 4

So we have that when x = 4, we must have y = 3.

3 = a*(4^2 -8*4 + 12) = a*-4

a = -3/4.

Our quadratic equation is:

y = (-3/4)*(x - 2)*(x-6)

User Tnwei
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