What is the maximum value and/or minimum value of the quadratic function f(x)= - 5x^2-10x6??

## what is the maximum value or minimum value of the quadratic function f(x)= -5x^2-10x-6?

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# 3 Answers

Hi Kevin;

f(x)= -5x

^{2}-10x^{6}The fact that both x values are to an even exponential (2 and 6), means that the results of any value of x will be positive, whether or not x is negative.

Henceforth, 0 is the minimum value of the results of x...

f(x)=-5(0

^{2})-10(0^{6})f(x)=0

However, both the 5x

^{2}and 10x^{6}are being subtracted. Henceforth, the results of the whole equation will always be negative except when x=0.0 is

**maximum**value of f(x), while 0 is the**minimum**value of x to either exponential.There is no

**minimum**value of f(x) because there is nothing else restricting the value of x.I think Nataliya's method is the best method outside of using derivatives, but just in case you are interested...

f(x) = -5x^2-10x-6

f'(x) = 2(-5) x

^{2-1 }- (1)10x^{1-1}+ 0f'(x) = -10x - 10

Set f'(x) = -10x - 10 = 0

-10x = 10

x = -1

f(-1) = -5(-1)

^{2}- 10(-1) - 6 = -5(1) + 10 - 6 = -5 + 4 = -1(-1, -1) is the vertex of a downward opening parabola, so y=-1 a maximum value.

Because the leading coefficient is negative the given quadratic function has maximum at vertex and has no minimum.

x-coordinate of the vertex is (- b/2a)

x = - (-10) / 2(- 5) = - 1

y = - 5 (- 1)

x-coordinate of the vertex is (- b/2a)

x = - (-10) / 2(- 5) = - 1

y = - 5 (- 1)

^{2}- 10(- 1) - 6 = - 1*max*_{f(x)}= - 1
## Comments

I meant student posted quadratic function f(x) = -5x^2 - 10x - 6

but you gave the answer for 6th power polynomial function max f(x) = 0 with min x = 0 , but there is no minimum for "x" whatsoever for those functions.

x-6x6-6and f(x)= - 5x^2-10x6, within the context of his other questions, you will see that the latter is consistent. He is not yet studying parabolas, lines of such symmetry, vertexes, etc.Comment