What are the zeros of this f?

Question

f(x)=6x^2+7x-24.

And what is the minimum or maximum value of f(x)?

Answer 1

6x^2+7x-24=6x^2+16x-9x-24=0
2x(3x+8)-3(3x+8)=0
(3x+8)(2x-3)=0
the zeros are -8/3 and 3/2

Answer 2

6x^2 + 7x - 24
factors to (2x-3) (3x+8)

So where 2x-3 = 0 and 3x+8 = 0 you'll cross the x axis
x=3/2 and x = -8/3 would be where y=0

The maximum value of f(x) would be infinity,
in this equation, the first derivitave 12x+7 = 0 would
tell you where the slope of the curve was 0 and give
you the minimum. 12x+7 =0, so x= -7 / 12,
plug x into the equation and you'll get about -26.04 as the y minimum.

Answer 3

To find the zeros replace each variable with zero. To find zeros of x make y=0 and vice versa.

The maximum or minimum value can be determined using the first derivative test...or you can graph it and cheat.

Answer 4

zeros are found using the quadratic formula in this case... you can figure that out for yourself.

min and max:

derive f wrt x

so,

f' = 12x + 7
f'' = 12

because f'' is a positive we know that this is a maximum.

set f' = 0, so we know that at x = -7/12 there is a maximum...

so, max value is

f(-7/12) = 6*(-7/12)^2+7*(-7/12) - 24

calculate that yourself. there is only one peak in this case.

Answer 5

f'(x)=12x+7
f'(x)=0 when 12x+7=0 meaning when x= -7/12

to find max min we difrentiate the function again, (12x+7)':
f''(x)=12 and 12 is > 0 therefore when x= -7/12 the function is at minimum, ther is no max point.

Answer 6

6x^2+7x-24=6x^2+16x-9x-24=0
2x(3x+8)-3(3x+8)=0
(3x+8)(2x-3)=0

the answer is -8/3 and 3/2

Answer 7

by "zeros", people are taking that to mean the "roots" of the equation, that is when y=0

Answer 8

Plug your formula into a graphing calculator and use it to find your zeros.