Question

unit 2 lesson 7: group work

1. rewrite in standard form: (x + 2)(x + 3)(x - 1)

then, state the degree, x - intercepts, and y - intercept.

1. rewrite in standard form: (x + 5)(x^2 + 2x + 3)
2. which points are the x - intercepts? which are relative maximums and minimums?

Explanation:

1.

Step1: Expand first two factors

$$(x + 2)(x + 3)=x^{2}+3x+2x + 6=x^{2}+5x + 6$$

Step2: Multiply by third factor

$$(x^{2}+5x + 6)(x - 1)=x^{3}-x^{2}+5x^{2}-5x+6x - 6=x^{3}+4x^{2}+x - 6$$
The degree is 3 (highest - power of x).
To find x - intercepts, set $y = 0$:
$$(x + 2)(x + 3)(x - 1)=0$$
$x=-2,x=-3,x = 1$ are the x - intercepts.
To find y - intercept, set $x = 0$:
$$y=(0 + 2)(0 + 3)(0 - 1)=-6$$

Step1: Use distributive property

$$(x + 5)(x^{2}+2x + 3)=x(x^{2}+2x + 3)+5(x^{2}+2x + 3)$$

Step2: Expand further

$$x^{3}+2x^{2}+3x+5x^{2}+10x + 15=x^{3}+7x^{2}+13x + 15$$
The degree is 3 (highest - power of x).
To find x - intercepts, we need to solve $x^{3}+7x^{2}+13x + 15 = 0$. This is a cubic equation. One way is to try some simple values of x (rational - root theorem). By trying $x=-3$:

$$(-3)^{3}+7(-3)^{2}+13(-3)+15=-27 + 63-39 + 15=12 eq0$$

To find y - intercept, set $x = 0$:
$$y=(0 + 5)(0^{2}+2\times0 + 3)=15$$

x - intercepts are the points where the graph crosses the x - axis, i.e., where $y = 0$. From the graph, the x - intercepts are points A, C, and G.
Relative maximum is a point where the function changes from increasing to decreasing. Point B is a relative maximum.
Relative minimum is a point where the function changes from decreasing to increasing. Points D and F are relative minima.

Answer:

Standard form: $x^{3}+4x^{2}+x - 6$; Degree: 3; x - intercepts: $x=-2,x=-3,x = 1$; y - intercept: $y=-6$

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