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 $(0, -1)$.

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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.