# The Absolute maximum of f(x)

#### jpratt

Consider the function f(x)=5x^2+8x-5. The absolute maximum of f(x) is____

so far i took the derivative of the given and got this 10x+8, from that i got -4/5.
so what to do from this to get the answer for the absolute maximum(Thinking)

#### pickslides

MHF Helper
so far i took the derivative of the given and got this 10x+8, from that i got -4/5.
That looks like the minimum to me.

#### Prove It

MHF Helper
Consider the function f(x)=5x^2+8x-5. The absolute maximum of f(x) is____

so far i took the derivative of the given and got this 10x+8, from that i got -4/5.
so what to do from this to get the answer for the absolute maximum(Thinking)
Since your $$\displaystyle a$$ value is positive, your quadratic will have a minimum at the turning point.

So that means that the absolute maximum will be at one of the endpoints. If there are not any endpoints, then the maximum value is $$\displaystyle \infty$$.

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#### pickslides

MHF Helper
Since your $$\displaystyle a$$ value is $$\displaystyle 1$$,
Hayden do you mean $$\displaystyle a=5$$ as in $$\displaystyle y=ax^2+bx+c$$ ?

#### Prove It

MHF Helper
Hayden do you mean $$\displaystyle a=5$$ as in $$\displaystyle y=ax^2+bx+c$$ ?
Yes I do. I meant since $$\displaystyle a > 0$$.

#### jpratt

how did u get 5 and do i but the original equation in y = ax^2 + bx+c

#### 11rdc11

if the coefficent infront of x^2 is positive than you have a min and if it neg then you a max. So since 5 is positive you would have a min

#### HallsofIvy

MHF Helper
"how did u get 5 and do i but the original equation in y = ax^2 + bx+c "

YOU said that the function was y= 5x^2+ 8x- 5. Since the leading coefficient is positive this is a parabola that opens upward. It has no "maximum value".

You could also see this by completing the square: y= 5(x^2+ (8/5)x+ 16/25- 16/25)- 5= 5(x+ 4/5)^2- 16/5- 25/5= 5(x+ 4/5)^2- 41/25.

When x= -4/5, the squared term will be 0 and y(-4/5)= -41/25. Since a square is never negative, y is never less than -41/25 so that is the minimum value. There is no bound on how large y can be and so no maximum.