[tex]f(x)=-x^2+5x-4[/tex]
a)
[tex](k,-4)\in f(x)\ \therefore\ -4=-k^2+5k-4\ \therefore\\\\ 0=-k^2+5k\ \therefore\ 0=k(5-k)\ \therefore\ \boxed{k=0}\ \ \text{ou}\ \ \boxed{k=5}[/tex]
b)
[tex]\boxed{[f(x)]_{m\acute{a}x}=y_V}=-\dfrac{b^2-4ac}{4a}=-\dfrac{5^2-4(-1)(-4)}{4(-1)}=[/tex]
[tex]=\dfrac{25-16}{4}=\dfrac{9}{4}\ \therefore\ \boxed{[f(x)]_{m\acute{a}x}=\dfrac{9}{4}}[/tex]
[tex]\mathbb{D}=(-\infty,\infty)\ \therefore\ \boxed{\mathbb{D}=\mathbb{R}}[/tex]
[tex]\mathbb{I}\text{m}=(-\infty,y_V)\ \therefore\ \boxed{\mathbb{I}\text{m}=\bigg(-\infty,\dfrac{9}{4}\bigg)}[/tex]
