Find the coordinates of the vertex for the parabola defined by the given quadratic function f(x) = (x + 2)^2 + 2 a) (0,2) b) (2.0) c) (-2.2) d) (-2,-2) e) none

Respuesta :

Answer:

(-2,2)

Step-by-step explanation:

Let's find the answer.

Because a tangent line for a parabola function is equal to 0 only at its vertex then:

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

[tex]f'(x)=2*(x+2)[/tex]

[tex]f'(x)=2x+4[/tex] so then:

[tex]f'(x)=0[/tex] when

[tex]0=2x+4[/tex]

[tex]-2=x[/tex]

For x=-2 f(x) is:

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

[tex]f(1)=(-2+2)^{2}+2[/tex]

[tex]f(x)=2[/tex]

In conclusion, the vertex of the given parabola is (-2,2), so the answer is C. Although in your answer is reported as (-2.2) but I think was a typing mistake.