Question

What is the range with steps? f(x)=x^2+8x+8

Answer 1

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

Find vertex to x

`\\ \sf\longmapsto (-b)/(2a)`

`\\ \sf\longmapsto (-8)/(2(1))`

`\\ \sf\longmapsto (-8)/(2)`

`\\ \sf\longmapsto -4`

• Find value of function at -4

`\\ \sf\longmapsto f(-4)`

`\\ \sf\longmapsto (-4)^2+8(-4)+8`

`\\ \sf\longmapsto 16-32+8`

`\\ \sf\longmapsto -16+8`

`\\ \sf\longmapsto -8`

The range is

`\\ \sf\longmapsto (8,\infty)`

Answer 2

• [-8, ∞)

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, ∞)