Respuesta :

We can see that

[tex](x+2)^2=x^2+4x+4[/tex]

By comparing this expression with our quadratic function, we get

[tex]y\questeq x^2+4x+4-4-16[/tex]

where we added and substracted 4, which gives zero. Now, we can write

[tex]\begin{gathered} y=(x+2)^2-4-16 \\ y=(x+2)^2-20 \end{gathered}[/tex]

Now, the quadratic function in vertex form is given by

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

where the point (h,k) is the vertex. By comparing our last result and this expression, we can see that h=2 and k=-20. Then, the vertex is at point (2,-20).

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