Let's analyze the function \( f(x) = 2x^2 - x - 1 \) and answer similar questions based on it.

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

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

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

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

\[
f(2) = 2(2)^2 - 2 - 1 = 2(4) - 2 - 1 = 8 - 2 - 1 = 5
\]

So, \( f(2) = 5 \). The point \((2, 5)\) is on the graph of \( f(x) \).

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

\[
2x^2 - x - 1 = -2
\]

Add 2 to both sides:

\[
2x^2 - x + 1 = 0
\]

Now, use the quadratic formula:

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

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

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

\[
x = \frac{1 \pm \sqrt{1 - 8}}{4} = \frac{1 \pm \sqrt{-7}}{4}
\]

Since the discriminant (\( -7 \)) is negative, there are no real solutions, meaning the function \( f(x) = 2x^2 - x - 1 \) does not intersect the line \( f(x) = -2 \) at any real point.

### (d) What is the domain of \( f(x) \)?
The domain of a quadratic function is all real numbers. Therefore, the domain of \( f(x) = 2x^2 - x - 1 \) is \( (-\infty, \infty) \).

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

\[
2x^2 - x - 1 = 0
\]

Use the quadratic formula:

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

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

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

\[
x = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4}
\]

So:

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

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

### (f) List the \( y \)-intercept, if there is one, of the graph of \( f(x) \).
To find the \( y \)-intercept, set \( x = 0 \) and solve for \( f(0) \):

\[
f(0) = 2(0)^2 - 0 - 1 = -1
\]

The \( y \)-intercept is \( (0, -1) \).

---

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 \( f(x) = 2x^2 - x - 1 \)?
2. What is the axis of symmetry for the function \( f(x) = 2x^2 - x - 1 \)?
3. How do you find the maximum or minimum value of \( f(x) = 2x^2 - x - 1 \)?
4. How does the value of \( a \) in the quadratic function affect the direction of the parabola?
5. How would you graph the function \( f(x) = 2x^2 - x - 1 \)?

**Tip:** For quadratic equations, knowing how to complete the square can be a useful alternative to the quadratic formula for finding roots and the vertex.

Explore the quadratic function f(x) = 2x^2 - x - 1 through detailed analysis. Learn about its vertex, domain, range, intercepts, and how to solve related problems. Suitable for students in Grades 9-12.

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