f(x)=3x^2-x-2
​(a) Is the point (-1,2) on the graph of​ f?
​(b) If x​ = 2​, what is​ f(x)? What point is on the graph of​ f?
​(c) If​ f(x) = negative 2​, what is​ x? What​ point(s) is/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.

Let's address each part of the question one by one for the function \( f(x) = 3x^2 - x - 2 \).

### (a) Is the point \((-1, 2)\) on the graph of \( f \)?
To determine if the point \((-1, 2)\) is on the graph, substitute \( x = -1 \) into the function and see if \( f(-1) = 2 \):

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

Since \( f(-1) = 2 \), the point \((-1, 2)\) is indeed on the graph of \( f(x) \).

### (b) If \( x = 2 \), what is \( f(x) \)? What point is on the graph of \( f \)?
Substitute \( x = 2 \) into the function:

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

So, \( f(2) = 8 \). The point \((2, 8)\) is on the graph of \( f(x) \).

### (c) If \( f(x) = -2 \), what is \( x \)? What point(s) is/are on the graph of \( f \)?
Set \( f(x) = -2 \) and solve for \( x \):

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

Adding 2 to both sides:

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

Factor the equation:

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

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

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

### (d) What is the domain of \( f \)?
The domain of a quadratic function is all real numbers since there are no restrictions on the values that \( x \) can take. Thus, the domain of \( f(x) \) is \( (-\infty, \infty) \).

### (e) List the \( x \)-intercept(s), if any, of the graph of \( f \).
To find the \( x \)-intercepts, set \( f(x) = 0 \) and solve for \( x \):

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

Use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 3 \), \( b = -1 \), and \( c = -2 \):

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-2)}}{2(3)}
\]

\[
x = \frac{1 \pm \sqrt{1 + 24}}{6} = \frac{1 \pm \sqrt{25}}{6} = \frac{1 \pm 5}{6}
\]

So:

\[
x = \frac{6}{6} = 1 \quad \text{or} \quad x = \frac{-4}{6} = -\frac{2}{3}
\]

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 \( f \).
To find the \( y \)-intercept, set \( x = 0 \) and solve for \( f(0) \):

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

The \( y \)-intercept is \( (0, -2) \).

---

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How do you determine the vertex of the quadratic function \( f(x) = 3x^2 - x - 2 \)?
2. What is the axis of symmetry for the function \( f(x) = 3x^2 - x - 2 \)?
3. How do you graph the function \( f(x) = 3x^2 - x - 2 \)?
4. What are the maximum and minimum values of the function \( f(x) = 3x^2 - x - 2 \)?
5. How does changing the coefficients of \( x^2 \), \( x \), and the constant term affect the shape and position of the parabola?

**Tip:** When solving quadratic equations, always check if they can be factored easily before using the quadratic formula—it can save time!

Explore the quadratic function f(x) = 3x^2 - x - 2 with detailed solutions to specific queries: (a) Is the point (-1, 2) on the graph? (b) Find f(2) and its corresponding point. (c) Solve for x when f(x) = -2 and identify the points on the graph. (d) Determine the domain of f. (e) List x-intercepts and (f) y-intercept of the graph of f. Suitable for Grades 9-12.

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