Let f(x)= 300x-11x^2-9. Find the maximum value of f to four decimal places

Don't know how to do this can you work it out for me

f(x)=300x-11x^(2)-9

The maximum or minimum of a quadratic function occurs at x=-(b)/(2a). If a is negative, the maximum value of the function is f(-(b)/(2a)). If a is positive, the minimum value of the function is f(-(b)/(2a)).
fmax(x)=ax^(2)+bx+c occurs at x=-(b)/(2a)

Multiply 2 by -11 to get -22.
x=-(300)/(-22)

Remove the single term factors from the expression.
x=-(-(300)/(22))

Reduce the expression -(300)/(22) by removing a factor of 2 from the numerator and denominator.
x=-(-(150)/(11))

Multiply -1 by each term inside the parentheses.
x=(150)/(11)

Squaring a number is the same as multiplying the number by itself (150*150). In this case, 150 squared is 22500.
f((150)/(11))=(-11*(22500)/(11^(2)))+30…

Squaring a number is the same as multiplying the number by itself (11*11). In this case, 11 squared is 121.
f((150)/(11))=(-11*(22500)/(121))+300((…

Cancel out all the common terms in the polynomial -11*(22500)/(121).
f((150)/(11))=(-1*(22500)/(11))+300((15…

Multiply -1 by (22500)/(11) to get -(22500)/(11).
f((150)/(11))=(-(22500)/(11))+300((150)…

Multiply 300 by each term inside the parentheses.
f((150)/(11))=-(22500)/(11)+(45000)/(11…

Combine all similar expressions.
f((150)/(11))=(22500)/(11)-9

The maximum value of the function is at x=-(b)/(2a). This is a maximum value because a is less than 0.
Answer: f((150)/(11))=(22401)/(11)

Recall that the x-coordinate of the vertex of f(x) = ax^2 + bx + c is -b/(2a). When a < 0 (like it is here), the y-coordinate of the vertex is where the maximum occurs.

If we re-write f(x) in this form, we get:
f(x) = -11x^2 + 300x - 9.

So, a = -11, b = 300, and c = -9.

Then, the value of x that maximizes f(x) is:
x = -b/(2a)
= -300/[2(-11)]
= 150/11.

Therefore, the maximum value of f(x) is:
f(150/11) = 300(150/11) - 11(150/11)^2 - 9 ≈ 2036.4545.

I hope this helps!