[tex]$y=-(x-2) ^2+16$[/tex] is the vertex form.
The minimum value of the given function [tex]f(x)=x^2+4x+12[/tex] exists 16.
Vertex form of the quadratic function
The vertex form of a quadratic function is given by
[tex]$y=a(x-h)^ 2+k$[/tex]
where (h, k) exists the vertex of the parabola.
when written in vertex form
(h, k) is the vertex of the parabola and x = h is the axis of symmetry
the h represents a horizontal shift (how far left, or right the graph has shifted from x = 0.
the k represents a vertical shift (how far up, or down the graph has shifted from y= 0.
Now let's convert this [tex]$y=-x^ 2+4 x+12$[/tex] into vertex form
[tex]$y=-x ^2+4 x+12$[/tex]
[tex]$y-12=-x ^2+4 x$[/tex]
[tex]$y-12=-(x ^2-4 x)$[/tex]
[tex]$y-12=-(x ^2-4 x+4-4)$[/tex]
[tex]$y-12=-(x ^2-4 x+4)+4$[/tex]
[tex]$y-16=-(x^ 2-4 x+4)$[/tex]
[tex]$y-16=-(x-2)^ 2$[/tex]
[tex]$y=-(x-2) ^2+16$[/tex] is the vertex form.
The minimum value of f(x) exists at the y value of the vertex equation.
That is when x = 2 then value of the function exists
[tex]$f(x)=- (2 - 2)^{2}+16= 16$[/tex]
[tex]$f(x)=x^{2}+4x+12 $[/tex] is 16
Thus, the minimum value of the given function [tex]f(x)=x^2+4x+12[/tex] exists 16.
To learn more about the vertex form of the quadratic function
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