Question

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

Answer 1

Answer:

A

Step-by-step 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

$x_{vertex}$ = - $\frac{b}{2a}$

y = x² - $\frac{5}{3}$ x + $\frac{31}{36}$ ← is in standard form

with a = 1 and b = - $\frac{5}{3}$ , thus

$\frac{x}{vertex}$ = - $\frac{-\frac{5}{3} }{2}$ = $\frac{5}{6}$

Substitute this value into y

y = ($\frac{5}{6}$ )² - $\frac{5}{3}$ ($\frac{5}{6}$ ) + $\frac{31}{36}$

   = $\frac{25}{36}$ - $\frac{25}{18}$ + $\frac{31}{36}$ = $\frac{1}{6}$

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

minimum value = $\frac{1}{6}$ → A