Answer:
[tex]\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}[/tex]
Step-by-step explanation:
As the function is
[tex]x^{2} +2x-8[/tex]
As we know that domain is the set of input values for which the function is defined.
Therefore,
[tex]\mathrm{Domain\:of\:}\:x^2+2x-8\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:<x<\infty \\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:\infty \:\right)\end{bmatrix}[/tex]
And range is defined as:
The set of the dependent variable for which the function is defined.
As
[tex]\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)[/tex]
[tex]\mathrm{If}\:a<0\:\mathrm{the\:range\:is}\:f\left(x\right)\le \:y_v[/tex]
[tex]\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v[/tex]
[tex]a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(-1,\:-9\right)[/tex]
[tex]f\left(x\right)\ge \:-9[/tex]
Therefore,
[tex]\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}[/tex]
The graph is also attached below.
Keywords: graph, domain, range
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