Find the maximum of $-2x^{2}+6x-3$ Question
Find the maximum of $-2x^{2}+6x-3$
My thoughts
I was wondering If I was supposed to format it as in this kind of format: $f(x)=k(x-a)^{2}+b$? so that I can find the coordinates or If i was supposed to factor it?
 A: We can use the oft-used "completing the square": $$f (x)=-2x^2+6x-3=-2 [x^2-3x+\frac {3}{2}] $$ $$\Rightarrow f (x)=-2 [(x-\frac {3}{2})^2-\frac {9}{4}+\frac {6}{4}] $$ Can you proceed now??
A: f(x) = $-2 (x^2 - 3x) - 3$
= $-2(x^2 - 3x + \frac94) - 3 + \frac92$
= $-2(x - \frac32)^2 + \frac{3}{2}$
Hope you can take it further.
A: As for any quadratic polynomial $ax^2+bx+c$, its extremum is obtained at $x_0=-\dfrac b{2a}$ and its value is $-\dfrac{\Delta}{4a}$. It is a minimum if $a>0$, a maximum if $a<0$.
Applying this result, we have a minimum at $x_0=\frac 32$ and this minimum is equal to
$$\frac{36-24}{8}=\frac32.$$
A: An alternative method is to calculate your function's derivative, and set it equal zero.
$$f(x)=-2x^2+6x-3$$
Differentiating:
$$f'(x)=-4x+6$$
Set $f'(x)=0$:
$$0=-4x+6$$
$$x=\frac{3}{2}$$
Knowing this, you can substitute this $x$ value into your original function and determine the value of the maximum.
To verify it is a maximum point, you can evaluate the second derivative $f''(x)$ at $x=\frac{3}{2}$:
$$f''(x)=-4$$
$$f''\left(\frac{3}{2}\right)=-4$$
Since $f''\left(\frac{3}{2}\right)<0$, it is concave down, thus it can be concluded that it is a local maximum.
A: Your idea is correct, and there are two ways of trying to solve the problem.



*

*The so-called "completing the square". In this method, you first try to cover the $k$ value, which is easy since it is the same as the scalar next to $x$, i.e. $2$. So you transform
$$-2x+6x-3$$
into $$-2(x^2-3x+\frac32)$$


Then, you take care of $a$, which must satisfy $2a=3$ because $2a$ is the coefficient next to $x$. So you now have $$-2((x-\frac32)^2 + b)$$
After this, you can either already see that the maximum is acchieved at $x=\frac32$, or you can actually calculate $b$.



*You could also just setup your equations, so you get


$$-2x^2+6x-3 = k(x-a)^2+b=kx^2 -2akx + ka^2 +b$$
and from that you get three equations:
$$-2=k\\
6=-2ak\\
-3=ka^2+b$$
and solve for $k,a,b$ (prefferably in that order) from there, so:


*

*$k=-2$

*$6=-2a\cdot (-2)$ meaning $a=\frac{6}{4}=\frac32$

*$-3=-2\cdot \left(\frac32\right)^2 + b$ meaning $b=-3+\frac92=\frac{3}{2}$

