Search 73,324 tutors
FIND TUTORS
Ask a question
0 0

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

Tutors, please sign in to answer this question.

3 Answers

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.
 
 

Comments

Vivian, for your function there are no minimum and no maximum for "x"
Hi Nataliya;
I do not understand your comment.
HAPPY THANKSGIVING!
Thank you, Happy Thanksgiving to you too. :)
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.
The first time the student entered the formula it was...
f(x)= -5x^2-10x-6
The second time it was...
f(x)= - 5x^2-10x6
The latter made more sense to me.
One more detail...
Kevin from Mesquite, Texas asks many good questions on Wyzant.  If you place this question and its two different presentations, f(x)= -5x^2-10x-6 and 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.
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) x2-1 - (1)10x1-1 + 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.
 
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

maxf(x) = - 1