Question

Find the maximum/minimum value of the function y = x2 - (5/3)x + 31/36.

A. 1/6

B. 5/6

C. 25/36

D. 10

Answer 1

A

Explanation:

Given a parabola in standard form, y = ax² + bx + c ( a ≠ 0 ), then

minimum/ maximum value is the y- coordinate of the vertex.

The x- coordinate of the vertex is

`x_(vertex)` = -
`(b)/(2a)`

y = x² -
`(5)/(3)` x +
`(31)/(36)` ← is in standard form

with a = 1 and b = -
`(5)/(3)` , thus

`(x)/(vertex)` = -
`(-(5)/(3) )/(2)` =
`(5)/(6)`

Substitute this value into y

y = (
`(5)/(6)` )² -
`(5)/(3)` (
`(5)/(6)` ) +
`(31)/(36)`

=
`(25)/(36)` -
`(25)/(18)` +
`(31)/(36)` =
`(1)/(6)`

Since a > 1 then the vertex is a minimum, thus

minimum value =
`(1)/(6)` → A