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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?

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    You can't formati it as $k(x-a) + b$ because the original expression is quadratic in $x$, while $k(x-a) + b$ is linear. – 2017-01-18
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    Ok i fixed it.. – 2017-01-18
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    Your function actually equals $-(\sqrt{2}x -\frac{3}{2}\sqrt{2})^2 + \frac{3}{2}$. – 2017-01-18

5 Answers 5

5

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??

2

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.

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    Notice that $-2(x^2 - 3x + \frac{9}{4}) - 3 - \frac{9}{2} \neq -2(x^2 - 3x) - 3$. By adding $9/4$ inside the parentheses, you have added $-2(9/4) = -9/2$ to $-2(x^2 - 3x) - 3$. To balance the equation, you must *add* $9/2$ outside the parentheses. You should have $f(x) = -2(x^2 - 3x + \frac{9}{4}) - 3 + \frac{9}{2}$. – 2017-05-19
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    Yes sign mistake. Thank you for noticing. – 2017-05-19
2

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 $$\frac{36-24}{8}=\frac32.$$

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    You're right: it's safer, but I think people should also know this formula exists. Thanks for pointing the error! – 2017-01-18
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    Alternatively just plug $\frac 32$ into the formula for $f(x)$ – 2017-01-18
2

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.

1

Your idea is correct, and there are two ways of trying to solve the problem.

  1. 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 $b$.

  1. 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 $b=-3+\frac92=\frac{3}{2}$