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krok68 [10]
3 years ago
14

Graph y=x^3 + 6x^2 + 8x and describe the end behavior

Mathematics
2 answers:
Nikitich [7]3 years ago
8 0
I can't really graph this for you but here's some points.

(X, Y):  (-4, 0), (-3, 3), (-2, 0), (-1, -3), (0,0)

Here's the end behavior:

x⇒∞, y⇒∞
x⇒-∞, y⇒-∞
shepuryov [24]3 years ago
8 0

Given function: y=x^3 + 6x^2 + 8x

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) = 0

Factoring quadratic x^2 +6x +8, we get

x(x+2)(x+4) =0

Applying zeros product rule, we get

x =0

x+2 = 0  => x = -2

x+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:

<h2>x⇒∞, y⇒∞</h2><h2>x⇒-∞, y⇒-∞ </h2>

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Find the distance d(P1, P2) between the points P1 and P2. P1 = (-5,3) P2 = (2,4) d(P1,P2) = (Simplify your answer. Type an exact
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Answer:

distance between two points is

5\sqrt{2}

Step-by-step explanation:

the distance d(P1, P2) between the points P1 and P2

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