Respuesta :

cerverusdante

Answer:

(-6, -36)

Step-by-step explanation:

The vertex [tex](h,k)[/tex] of a function of the form [tex]f(x)=ax^2+bx+c[/tex] is given by the formula:

[tex]h=\frac{-b}{2a}[/tex]

[tex]k=f(h)[/tex] in other words, we find h and then evaluate function at h to find k.

We know from our function that [tex]a=1[/tex], [tex]b=12[/tex].

Replacing values

[tex]h=\frac{-12}{2(1)}[/tex]

[tex]h=-\frac{12}{2}[/tex]

[tex]h=-6[/tex]

Now we can evaluate our function at -6 to find k:

[tex]k=f(h)=f(-6)[/tex]

[tex]k=(-6)^2+12(-6)[/tex]

[tex]k=36-72[/tex]

[tex]k=-36[/tex]

We can conclude that the vertex (h, k) of our function is (-6, -36)

kudzordzifrancis

Answer:

(-6,-36)

Step-by-step explanation:

The given function is

[tex]f(x)=x^2+12x[/tex]

We complete the square to write this function in the vertex form.

Add and subtract the square of half the coefficient of x.

[tex]f(x)=x^2+12x+6^2-6^2[/tex]

[tex]f(x)=x^2+12x+36-36[/tex]

The first three terms is now a perfect square trionomial

[tex]f(x)=(x+6)^2-36[/tex]

Or

[tex]f(x)=(x--6)^2-36[/tex]

The function is now in the form:

[tex]f(x)=a(x-h)^2+k[/tex]

where h=-6 and k=-36

The vertex is therefore (h,k)=(-6,-36)

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