Graph y=x^3 + 6x^2 + 8x and describe the end behavior
Question
Answer:
Given function: [tex]y=x^3 + 6x^2 + 8x[/tex]In order to graph it, let us find some coordinates for the given function to plot on graph.Let us find the x-intercepts first by setting given function equal to 0.x^3 + 6x^2 + 8x =0.Factoring out x.x(x^2+6x+8) = 0Factoring quadratic x^2 +6x +8, we get x(x+2)(x+4) =0Applying zeros product rule, we get x =0 x+2 = 0 => x = -2x+4 =0 => x = -4.Therefore, we got x-intercepts (0,0), (-2,0) and (-4,0).Because degree is 3 and leading coefficient a positive number, the graph would go down on the left and go up on the right.From the graph we can see end behaviour: x⇒∞, y⇒∞x⇒-∞, y⇒-∞
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