Answer the questions about the following function.
f left parenthesis x right parenthesis equals 3 x squared minus x minus 2
​(a) Is the point left parenthesis 2 comma 8 right parenthesis on the graph of​ f?
​(b) If x​ = negative 1​, what is​ f(x)? What point is on the graph of​ f?
​(c) If​ f(x) = negative 2​, what is​ x? What​ point(s) are on the graph of​ f?
​(d) What is the domain of​ f?
​(e) List the​ x-intercept(s), if​ any, of the graph of f.
​(f) List the​ y-intercept, if there is​ one, of the graph of f.

Given the function \( f(x) = 3x^2 - x - 2 \), let's answer each question step by step:

### (a) Is the point \( (2, 8) \) on the graph of \( f \)?

To check if the point \( (2, 8) \) is on the graph, we need to evaluate \( f(2) \) and see if the result equals 8.

\[
f(2) = 3(2)^2 - (2) - 2 = 3(4) - 2 - 2 = 12 - 2 - 2 = 8
\]

Since \( f(2) = 8 \), the point \( (2, 8) \) is indeed on the graph of \( f \).

### (b) If \( x = -1 \), what is \( f(x) \)? What point is on the graph of \( f \)?

We need to calculate \( f(-1) \):

\[
f(-1) = 3(-1)^2 - (-1) - 2 = 3(1) + 1 - 2 = 3 + 1 - 2 = 2
\]

So, \( f(-1) = 2 \), which means the point \( (-1, 2) \) is on the graph of \( f \).

### (c) If \( f(x) = -2 \), what is \( x \)? What point(s) are on the graph of \( f \)?

We solve for \( x \) when \( f(x) = -2 \):

\[
3x^2 - x - 2 = -2
\]

Add 2 to both sides:

\[
3x^2 - x = 0
\]

Factor the equation:

\[
x(3x - 1) = 0
\]

So, \( x = 0 \) or \( 3x - 1 = 0 \), which gives \( x = \frac{1}{3} \).

Thus, the points \( (0, -2) \) and \( \left( \frac{1}{3}, -2 \right) \) are on the graph of \( f \).

### (d) What is the domain of \( f \)?

The function \( f(x) = 3x^2 - x - 2 \) is a polynomial, and polynomials are defined for all real numbers. Therefore, the domain of \( f \) is:

\[
\text{Domain of } f = (-\infty, \infty)
\]

### (e) List the \( x \)-intercept(s), if any, of the graph of \( f \).

The \( x \)-intercepts occur when \( f(x) = 0 \):

\[
3x^2 - x - 2 = 0
\]

We solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-2)}}{2(3)} = \frac{1 \pm \sqrt{1 + 24}}{6} = \frac{1 \pm \sqrt{25}}{6} = \frac{1 \pm 5}{6}
\]

This gives two solutions:

\[
x = \frac{1 + 5}{6} = 1 \quad \text{and} \quad x = \frac{1 - 5}{6} = -\frac{2}{3}
\]

Therefore, the \( x \)-intercepts are \( (1, 0) \) and \( \left( -\frac{2}{3}, 0 \right) \).

### (f) List the \( y \)-intercept, if there is one, of the graph of \( f \).

The \( y \)-intercept occurs when \( x = 0 \):

\[
f(0) = 3(0)^2 - 0 - 2 = -2
\]

So, the \( y \)-intercept is \( (0, -2) \).

---

Do you want more details on any step? Here are five related questions you might want to explore:
1. What is the vertex of the function \( f(x) = 3x^2 - x - 2 \)?
2. How would you find the axis of symmetry for this quadratic function?
3. What is the range of the function \( f(x) = 3x^2 - x - 2 \)?
4. Can you sketch the graph of the function using the points we found?
5. What is the behavior of \( f(x) \) as \( x \to \infty \) and \( x \to -\infty \)?

**Tip**: For quadratic functions, the vertex can be found using the formula \( x = -\frac{b}{2a} \), where \( a \) and \( b \) come from the standard form \( ax^2 + bx + c \).

Answer the questions about the following function. f(x) = 3x^2 - x - 2. (a) Is the point (2, 8) on the graph of f? (b) If x = -1, what is f(x)? What point is on the graph of f? (c) If f(x) = -2, what is x? What point(s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercept(s), if any, of the graph of f. (f) List the y-intercept, if there is one, of the graph of f.

Explore step-by-step solutions to questions about the quadratic function f(x) = 3x² - x - 2, including checking points on the graph, finding intercepts, solving for x, and determining the domain. Suitable for students in Grades 9-11.

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