Question

3.

Answer 1

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.

$f(x)=-{x}^{2}-3x+6$

First: let us take out the common factor: $-1$, since we remember that the canonical form is characterized as follows:

$\boxed{f(x)=a{(x-h)}^{2}+k$

Then, it remains:

$f(x)=-1({x}^{2}+3x-6)$

Then: the coefficient of the variable $x$ We divided it between $2$, And we square it (they will be one positive and one negative). In our case:

$\dfrac{3}{2}\rightarrow {\dfrac{3}{2}}^{2}$

$\frac{9}{4}[\tex]We apply it to the function:[tex]f(x)=-1({x}^{2}+3x-6+ \boldsymbol{\dfrac{9}{4}}- \boldsymbol{\dfrac{9}{4}})$

Let's accommodate terms to make it easier:

$f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-6-\frac{9}{4}$

$-6$ Can be written as $-\frac{24}{4}$:

$f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-\frac{24}{4}-\frac{9}{4}$

$f(x)=-1(\underline{{x}^{2}+3x+\frac{9}{4}}-\frac{33}{4}$

Now, what is underlined is our perfect square trinomial, let us recall its form:

$\boxed{{a}^{2}+2ab+{b}^{2}}$

Applying the same principle we are left:

$f(x)=-1[{(x+\frac{3}{2})}^{2}-\frac{33}{4})$

Applying distributive property we get:

$\boxed{\boxed{\boxed{f(x)=-{(x+\frac{3}{2})}^{2}+\frac{33}{4}}}}$

Therefore it will have its vertex in: $(-1.5,\:8.25)$

The axis of symmetry is a straight line that makes the function to be projected being $2$, for this you need some reference point, for the parabola you need the coordinate in $x$ of the vertex.

For which the axis of symmetry is $-1.5$.

I love you so much !