Question

The minimum/maximum value of the function y = a(x − 2)(x − 1) occurs at x = d, what is the value of d?

Answer 1

Answer:

$d=\frac{3}{2}=1.5$

Step-by-step explanation:

We have the function:

$y=a(x-2)(x-1)$

And we want to find x=d for which the minimum/maximum value will occur.

Notice that our function is a quadratic in factored form.

Remember that the minimum/maximum value always occurs at the vertex point.

And remember that the x-coordinate of the vertex is the axis of symmetry.

Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.

Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.

Our factors are (x-2) and (x-1).

Therefore, our roots/zeros are x=1, 2.

So, the average of them are:

$d=\frac{1+2}{2}=3/2=1.5$

Therefore, regardless of the value of a, the minimum/maximum value will occur at x=d=1.5.

Alternative Method:

Of course, we can also expand to confirm our answer. So:

$y=a(x^2-2x-x+2)\\y=a(x^2-3x+2)\\y=ax^2-3ax+2a$

The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.

And the formula for the vertex is:

$x=-\frac{b}{2a}$

So, we will substitute -3a for b and a for a. This yields:

$x=-\frac{-3a}{2a}=\frac{3}{2}=1.5$

Confirming our answer.