Hello Papi :D
To solve your problem we must first know what parable it is. In this case, since it is a quadratic function, we can only use two types of parabola: the vertical up and the vertical down.
Why can't the horizontal line be used?
It is not used because it breaks the function definition: for every value of [tex]x[/tex] there is one of [tex]y[/tex], in this type of parabola for every value of [tex]x[/tex] there is 2 of [tex]y[/tex].
Continuing: in our case it is a downward parabola, its form is:
[tex]{x}^{2}=-4py[/tex]
Where [tex]p[/tex] is the parameter.
Let's start with the function:
[tex]f(x)=\frac{1}{4}{(x+3)}^{2}+2[/tex]
From knowledge acquired in the course of quadratic function we realize that the canonical form is presented. We found it this way:
[tex]\boxed{f(x)=a{(x-h)}^{2}+k}[/tex]
Where you will find the center in [tex](h,k)[/tex], Being the coordinates of the same [tex]\boldsymbol{(-3,2)}[/tex].
For the function we find some cartesian coordinate, whatever you want, in this case I assigned the value of [tex]x=0[/tex], so:
[tex]f(0)=-\frac{1}{4}{(0+3)}^{2}+2[/tex]
[tex]f(0)=-\frac{9}{4}+\frac{8}{4}[/tex]
[tex]f(0)=-\frac{1}{4}[/tex]
The coordinate is: [tex](0,-\frac{1}{4})[/tex], This we replace in [tex]{x}^{2}=-4py[/tex], But as we have already seen, it is outside the origin, so it changes:
[tex]{(x-h)}^{2}=-4p(y-k)[/tex]
[tex]{(x-3)}^{2}=-4p(y-2)[/tex]
Substituting:
[tex]{(0-3)}^{2}=-4p(-\frac{1}{4}-\frac{8}{4})[/tex]
[tex]9=-4p(-\frac{9}{4})[/tex]
[tex]9=9p[/tex]
[tex]\boxed{\boldsymbol{p=1}}[/tex]
Remember that [tex]p[/tex] It is the distance between the focus and the vertex, and also between the vertex and the straight line.
Taking the distance we can find the focus, we have the center (or vertex) which is
[tex](-3,2)[/tex] and the distance, which is [tex]1[/tex], being in the same coordinate of the abscissa axis (since the focus lies below the vertex), then what we do is to subtract from the vertex ordinate one unit to obtain the focus:
[tex]F(-3,2-1)[/tex]
[tex]\mathbb{ANSWER}\Rightarrow \boxed{\boxed{\boxed{F(-3,1)}}}[/tex]
Greetings and long live all the anime of the world.
Especially Gintama, I love Gintama :)
