Maximum of a Fraction

To find the maximum value of the fraction, we find the minimum value of the quadratic in the denominator.
Let,
$f(x) = x^2-2x+a$
$f'(x) = 2x-2 = 0$
$\Rightarrow x = 1$

Now, $\displaystyle x=1$ could be the maximum or the minimum. We can confirm that it is the minimum by finding the second derivative of the function or another way is that we can see the concavity of the function (in this case, it is concave upwards, hence it would give a minimum value)

Substituting $\displaystyle x=1$ in $\displaystyle f(x)$ , we get,
$f(1) = a-1$

Putting the value in the original fraction,
The maximum value of the fraction is,

$\frac{27}{a-1} = 3$

$\Rightarrow\boxed{a=10}$

$$ The value of $\dfrac{27}{x^2-2x+a}$ will be maximum if the value of $x^2-2x+a$ is minimum. Therefore $$ $\begin{aligned} x^2-2x+a&=\frac{27}{3}\\x^2-2x+a-9&=0. \end{aligned}$ $$ In general, quadratic equation $\,f(x)=ax^2+bx+c$ , for $a>0$ , will have a minimum value at $x=-\dfrac{b}{2a}$ . Thus $$ $x=-\frac{(-2)}{2(1)}=1$ and $\begin{aligned} x^2-2x+a-9&=0\\ 1^2-2(1)+a-9&=0\\ a&=\boxed{10} \end{aligned}$ $$ $\text{\# }\mathbb{Q.E.D.}\text{ \#}$

We can also write the denominator as $(x+1)^{2}+a-1$ Hence the minimum value of the denominator produces the maximum value of the function which is $\frac{27}{a-1}$ ....Since this value is 3...we have $a = 10$ ..

If the maximum value of $\dfrac{27}{x^2 - 2x + a}$ is 3, then the minimum of $x^2 - 2x + a$ is 9. The $y$ -coordinate of a vertex is given by $-D/4p$ in the quadratic equation $px^2 + qx + r=0$ and $D = q^2 - 4pr$ , also known as the discriminant. In this question, $D = 4 - 4a$ and the minimum is 9.

And so, finally, we are left with the equation $\dfrac{-(4 - 4a)}{4} = 9 \Rightarrow \boxed{a = 10}$