Question

Which of the following equations has a minimum value of (3,-10)? y = 2x2 + 40x + 203 y = x2 + 6x + 19 y = 2x2 − 12x + 8 y = -2x2 + 12x − 8

Answer 1

y = 2x² − 12x + 8

Explanation:

FIRST METHOD :

y = 2x² − 12x + 8

= (2x² − 12x) + 8

= 2 (x² − 6x) + 8

= 2 (x² − 6x + 9 − 9 ) + 8

= 2 (x² − 6x + 9) − 2×9 + 8

= 2 (x² − 6x + 9) − 18 + 8

= 2 (x² − 6x + 9) − 10

= 2 (x − 3)² − 10

Then ,the equation has a extremum value of (3,-10)

Since the number 2 in the equation y = = 2 (x − 3)² − 10 is greater than 0

(2 > 0) , the graph (parabola) opens upward

Therefore ,the extremum (3,-10) is a minimum.

SECOND METHOD :

the graph of a function of the form f(x) = ax² + bx + c

has an extremum at the point :

`\left( -(b)/(2a) ,f\left( -(b)/(2a) \right) \right)`

in the equation : f(x) = 2x² − 12x + 8

a = 2 ; b = -12 ; c = 8

Then

`-(b)/(2a) = -(-12)/(2 * 2) = 3`

Then

`f\left( -(b)/(2a) \right) = f(3) = 2(3)^2- 12(3) + 8 = 18 - 36 + 8 = -18 + 8 = -10`

the graph of a function f(x) = 2x² − 12x + 8

has an extremum at the point (3 , -10)

Since the parabola opens up ,then the extremum (3,-10) is a minimum.