Answer:
[tex]d=\frac{3}{2}=1.5[/tex]
Step-by-step explanation:
We have the function:
[tex]y=a(x-2)(x-1)[/tex]
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:
[tex]d=\frac{1+2}{2}=3/2=1.5[/tex]
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:
[tex]y=a(x^2-2x-x+2)\\y=a(x^2-3x+2)\\y=ax^2-3ax+2a[/tex]
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:
[tex]x=-\frac{b}{2a}[/tex]
So, we will substitute -3a for b and a for a. This yields:
[tex]x=-\frac{-3a}{2a}=\frac{3}{2}=1.5[/tex]
Confirming our answer.
