Respuesta :

mhanifa

Answer:

  • [-8, ∞)

Step-by-step explanation:

Given function:

  • f(x)=x² + 8x + 8

This is a quadratic function with positive leading coefficient, therefore it is an increasing function with its minimum at vertex. It has no maximum value.

The vertex is at x = -b/2a:

  • x = -8/2 = -4

The minimum value is:

  • f(-4) = (-4)² + 8(-4) + 8 = 16 - 32 + 8 = -8

The range of the function is:

  • [-8, ∞)

f(x)=x^2+8x+8

Find vertex to x

[tex]\\ \sf\longmapsto \dfrac{-b}{2a}[/tex]

[tex]\\ \sf\longmapsto \dfrac{-8}{2(1)}[/tex]

[tex]\\ \sf\longmapsto \dfrac{-8}{2}[/tex]

[tex]\\ \sf\longmapsto -4[/tex]

  • Find value of function at -4

[tex]\\ \sf\longmapsto f(-4)[/tex]

[tex]\\ \sf\longmapsto (-4)^2+8(-4)+8[/tex]

[tex]\\ \sf\longmapsto 16-32+8[/tex]

[tex]\\ \sf\longmapsto -16+8[/tex]

[tex]\\ \sf\longmapsto -8[/tex]

The range is

[tex]\\ \sf\longmapsto (8,\infty)[/tex]