Determine the linear function $f: R \rightarrow R$ There's a linear function $f$ whose correspondence rule is $f(x)=|ax^2-3ax+a-2|+ax^2-ax+3$. State the values of the parameter $a$ that fully define the function $f$.
According to the problem condition the function $ f (x) $ is linear by
so the coefficient that accompanies $ x ^ 2 $ must be equal to 0, this is
meets when:
\begin{align*}                                                               
  ax^2-3ax+a-2 & < 0\\                                                       
 a(x^2-3x) &<2-a \\                                                         
  \                                   
    a\underbrace{\left(x^2-3x+\frac{9}{4}\right)}_{                          
  \left(x-\frac{3}{2}\right)^2≥ 0} &<2-a +\frac{9}{4}a\\                     
  a \left(x-\frac{3}{2}\right)^2 &<2+\frac{5}{4}a\\                          
  0 & <2+\frac{5}{4}a \\                                                     
  -\frac{8}{5} & <a                                                          
\end{align*}

What else is missing to analyze?
 A: For any $a$, in order for the $x^2$ term to vanish, the inequality $ax^2-3ax+a-2\le0$ must hold for all values of $x$. Using your analysis, this is equivalent to
$$a\left(x-\frac32\right)^2\le2+\frac54a$$
If $a > 0$, there is some $x$ large/small enough such that LHS > RHS. Hence $a \le 0$.
Moreover, if $2 + \dfrac54 a < 0$, the above inequality fails for $x = \dfrac32$. Hence $a \ge -\dfrac 85$.
This gives the range $-\dfrac 8 5 \le a \le 0$; indeed in this range, LHS is nonpositive for all $x$ while RHS is nonnegative, so our initial inequality holds, and the function is linear.
A: If there's some $x_0$ such that $ax_0^2-3ax_0+a-2>0$, then there's some neighborhood around $x_0$ such that $\forall x \in (x_0-\delta, x_0+ \delta)$, $ax^2-3ax+a-2>0$, $f(x)$ is quadratic unless $a=0$. On the other hand if $a=0$ then $f(x)$ is indeed linear.
If $a\ne 0$, it follows that $ax^2-3ax+a-2 \le 0, \forall x \in \mathbb R$. Therefore $a<0$ and the discriminant is non-negative, i.e.
$$(-3a)^2 \le 4 a (a-2) \iff a(5a+8) \le 0 \iff -\frac{8}{5} \le a <0.$$
Therefore $-\frac{8}{5} \le a \le 0$.
