Hello Papi :D
I will solve the problem by applying the perfect square trinomial. In this way we obtain the canonical form. Another way would be to derive the function, but I don't know if you're familiar with it.
[tex]f(x)=-{x}^{2}-3x+6[/tex]
First: let us take out the common factor: [tex]-1[/tex], since we remember that the canonical form is characterized as follows:
[tex]\boxed{f(x)=a{(x-h)}^{2}+k[/tex]
Then, it remains:
[tex]f(x)=-1({x}^{2}+3x-6)[/tex]
Then: the coefficient of the variable [tex]x[/tex] We divided it between [tex]2[/tex], And we square it (they will be one positive and one negative). In our case:
[tex]\dfrac{3}{2}\rightarrow {\dfrac{3}{2}}^{2}[/tex]
[tex]\frac{9}{4}[\tex]
We apply it to the function:
[tex]f(x)=-1({x}^{2}+3x-6+ \boldsymbol{\dfrac{9}{4}}- \boldsymbol{\dfrac{9}{4}})[/tex]
Let's accommodate terms to make it easier:
[tex]f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-6-\frac{9}{4}[/tex]
[tex]-6[/tex] Can be written as [tex]-\frac{24}{4}[/tex]:
[tex]f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-\frac{24}{4}-\frac{9}{4}[/tex]
[tex]f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-\frac{33}{4}[/tex]
Now, what is underlined is our perfect square trinomial, let us recall its form:
[tex]\boxed{{a}^{2}+2ab+{b}^{2}}[/tex]
Applying the same principle we are left:
[tex]f(x)=-1[{(x+\frac{3}{2})}^{2}-\frac{33}{4})[/tex]
Applying distributive property we get:
[tex]\boxed{\boxed{\boxed{f(x)=-{(x+\frac{3}{2})}^{2}+\frac{33}{4}}}}[/tex]
Therefore it will have its vertex in: [tex](-1.5,\:8.25)[/tex]
The axis of symmetry is a straight line that makes the function to be projected being [tex]2[/tex], for this you need some reference point, for the parabola you need the coordinate in [tex]x[/tex] of the vertex.
For which the axis of symmetry is [tex]-1.5[/tex].
I love you so much !
