Determine the zeros and the max. Or min. Value for the function. 6x^2+7x-3Explain. Thanks

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Let `f(x)= 6x^2 + 7x -3` .

First we will find the zeros.

==> `x1 = (-7+sqrt(49+12))/(2*6)= (-7+11)/12 = 4/12 = 1/3 `

`==gt x2= (-7-11)/12 = -18/12 = -3/2`

`` Then the zeros are `1/3`  and `-3/2`

`` Now we need to find the extreme value (Min. or Max.)

First we note that the sign of the factor of `x^2`  is positive. Then, we know that the function is facing up. Hence, `f(x)`  has a minimum value.

Now we will determine the minimum value by differentiate f(x).

==> `f'(x)= 12x + 7` .

Now we will find the derivatives zero.

==> `12x+7= 0==gt x = -7/12`

`` Then the function has a minimum value at `x= -7/12`

`` ==>`f(-7/12)= 6(-7/12)^2 +7(-7/12) -3 `

`==gt f(-7/12)= 42/144- 49/12 - 3 = (7-98-36)/24 = -127/24 = -5.29`

`` ==> Then the minimum point is `(-7/12, -127/24)`

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