Question

Find the maximum value of the quadratic equation -6x²+3x+2 ?
please explain...
a) -19/8
b) -25/8
c) 25/8
d) 19/8
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Answer 1

Step-by-step explanation:

maximum value of a quadratic equation will be at x= (-b/2a)

so (-3/(2*-6)) = -3/-12 = 1/4

maximum value = -6(1/4)^2+3(1/4)+2

= -6(1/16)+3/4+2

= -3/8 +3/4 +2

= (-3+6+16)/8

= 19/8

so option d

Answer 2

Answer:

The maximum value of the given quadratic equation is equal to 19/8.

Therefore the option (d) is correct.

Step-by-step explanation:

The given expression:   $y=-6x^{2}+3x+2$              ............(1)

Differentiate eq.(1) w.r.t. 'x':

$\frac{dy}{dx}=\frac{d}{dx} (-6x^{2}+3x +2 )$

$\frac{dy}{dx}=-12x+3$                                                             ...............(2)

Now,  $\frac{dy}{dx} =0$

  $-12x+3=0$

      $12x=3\\x=\frac{1}{4}$

Further derivative of eq.(2),

$\frac{d^{2}x}{dy^{2}}=-12 < 0$    ⇒ the maximum value

If  $\frac{d^{2}y}{dx^{2}} > 0$   then minimum value

If  $\frac{d^{2}y}{dx^{2}} < 0$  then maximum value

Therefore put x=1/5 in eq.(1) to get maximum value:

$y_{max}=-6(\frac{1}{4})^{2} +3(\frac{1}{4} )+2$

$y_{max}=\frac{-6}{16} +\frac{3}{4} +2$

$y_{max}=\frac{-6+12+32}{16}$

$y_{max}=\frac{38}{16}$

$y_{max}=\frac{19}{8}$

Therefore the maximum value is 19/8.