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

Question

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

Answer 1

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 + 0

f'(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.

 

Answer 2

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)² - 10(- 1) - 6 = - 1

max_f(x) = - 1 

Answer 3

Hi Kevin;

f(x)= -5x²-10x⁶

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(0²)-10(0⁶)

f(x)=0

However, both the 5x² and 10x⁶ 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.