Question

The ranger of the function x2+2x-8is allá

Answer 1

Answer:

$\mathrm{Range\:of\:}x^2+2x-8:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-9\:\\ \:\mathrm{Interval\:Notation:}&\:[-9,\:\infty \:)\end{bmatrix}$

Step-by-step explanation:

As the function is

$x^{2} +2x-8$

As we know that domain is the set of input values for which the function is defined.

Therefore,

$\mathrm{Domain\:of\:}\:x^2+2x-8\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:$

And range is defined as:

The set of the dependent variable for which the function is defined.

As

$\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)$

$\mathrm{If}\:a$

$\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v$

$a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(-1,\:-9\right)$

$f\left(x\right)\ge \:-9$

Therefore,

$\mathrm{Range\:of\:}x^2+2x-8:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-9\:\\ \:\mathrm{Interval\:Notation:}&\:[-9,\:\infty \:)\end{bmatrix}$

The graph is also attached below.

Keywords: graph, domain, range

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