Answer:
[tex]-4x^{2} +11[/tex] is a parabola. The coefficient in front of the highest term (-4 in front of the [tex]x^{2}[/tex]) is negative, meaning both of the ends go towards negative infinity (point downward). So, as x approaches positive or negative infinity, y approaches negative infinity.
The end behavior of the function is being divergent for [tex]x \to \pm \infty[/tex].
According to theory of limits and differential calculus, all polynomial functions are continuous and differentiable function with no limits for [tex]x \to \pm \infty[/tex]. Therefore, we can conclude that function [tex]h(x) = -4\cdot x^{2}+11[/tex] diverges to [tex]-\infty[/tex] for [tex]x > 0[/tex] or [tex]x < 0[/tex].
That is,
[tex]x \to +\infty[/tex]
[tex]\lim_{x \to +\infty} -4\cdot x^{2} + 11 = -4 \cdot \lim_{x \to +\infty} x^{2} + \lim_{x \to +\infty} 11[/tex]
[tex]\lim_{x \to +\infty} -4\cdot x^{2} + 11 = N.E. + 11[/tex]
[tex]\lim_{x \to +\infty} -4\cdot x^{2} + 11 = N.E.[/tex]
[tex]x \to -\infty[/tex]
[tex]\lim_{x \to -\infty} -4\cdot x^{2} + 11 = -4 \cdot \lim_{x \to -\infty} x^{2} + \lim_{x \to -\infty} 11[/tex]
[tex]\lim_{x \to -\infty} -4\cdot x^{2} + 11 = N.E. + 11[/tex]
[tex]\lim_{x \to -\infty} -4\cdot x^{2} + 11 = N.E.[/tex]
Hence, the end behavior of the function is being divergent for [tex]x \to \pm \infty[/tex].
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