QUESTION IMAGE
Question
- given the function: $f(x)=2x(x + 3)(x - 2).$
a. find its x-intercepts and plot them.
b. determine its end behavior.
as $x \to +\infty$, $f(x) \to \underline{+\infty}$
as $x \to -\infty$, $f(x) \to \underline{-\infty}$
c. find its y-intercept.
d. make a table for values between and beyond the x-intercepts. then draw the graph of the function. be sure to scale and label your y-axis properly.
| $x$ | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
|---|
Step1: Find x-intercepts
Set $f(x)=0$, so $2x(x+3)(x-2)=0$. Solve for $x$:
$2x=0 \implies x=0$; $x+3=0 \implies x=-3$; $x-2=0 \implies x=2$
Step2: Confirm end behavior
Expand $f(x)$: $f(x)=2x(x^2+x-6)=2x^3+2x^2-12x$. Leading term $2x^3$:
As $x\to+\infty$, $2x^3\to+\infty$; as $x\to-\infty$, $2x^3\to-\infty$
Step3: Find y-intercept
Set $x=0$: $f(0)=2(0)(0+3)(0-2)=0$
Step4: Calculate table values
Substitute $x$ into $f(x)$:
- $x=-4$: $f(-4)=2(-4)(-4+3)(-4-2)=2(-4)(-1)(-6)=-48$
- $x=-3$: $f(-3)=2(-3)(0)(-5)=0$
- $x=-2$: $f(-2)=2(-2)(1)(-4)=16$
- $x=-1$: $f(-1)=2(-1)(2)(-3)=12$
- $x=0$: $f(0)=0$
- $x=1$: $f(1)=2(1)(4)(-1)=-8$
- $x=2$: $f(2)=2(2)(5)(0)=0$
- $x=3$: $f(3)=2(3)(6)(1)=36$
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a. x-intercepts: $x=-3$, $x=0$, $x=2$
b. As $x\to+\infty$, $f(x)\to+\infty$; As $x\to-\infty$, $f(x)\to-\infty$
c. y-intercept: $(0, 0)$
d. Completed table:
| $x$ | $-4$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
|---|
(Graph instructions: Plot the intercepts $(-3,0)$, $(0,0)$, $(2,0)$ and the table points, then draw a smooth cubic curve passing through all points, matching the end behavior.)