25. f(x)=2x^2-x-1 (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)=-1 , what is x? What point (s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercepts, 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 step by step.

### (a) Is the point $(-1, 2)$ on the graph of $f$?

To determine if the point $(-1, 2)$ is on the graph of $f(x) = 2x^2 - x - 1$, we need to check if $f(-1) = 2$.

Calculate $f(-1)$:
$$f(-1) = 2(-1)^2 - (-1) - 1$$
$$f(-1) = 2(1) + 1 - 1$$
$$f(-1) = 2 + 1 - 1$$
$$f(-1) = 2$$

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

### (b) If $x = -2$, what is $f(x)$? What point is on the graph of $f$?

Calculate $f(-2)$:
$$f(-2) = 2(-2)^2 - (-2) - 1$$
$$f(-2) = 2(4) + 2 - 1$$
$$f(-2) = 8 + 2 - 1$$
$$f(-2) = 9$$

So, when $x = -2$, $f(x) = 9$. The point $(-2, 9)$ is on the graph of $f$.

### (c) If $f(x) = -1$, what is $x$? What point(s) are on the graph of $f$?

We need to solve the equation $2x^2 - x - 1 = -1$:
$$2x^2 - x - 1 + 1 = 0$$
$$2x^2 - x = 0$$
$$x(2x - 1) = 0$$

This gives us two solutions:
$$x = 0 \quad \text{or} \quad 2x - 1 = 0$$
$$2x - 1 = 0 \implies x = \frac{1}{2}$$

So, the values of $x$ are $0$ and $\frac{1}{2}$. The points on the graph are $(0, -1)$ and $\left(\frac{1}{2}, -1\right)$.

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

The function $f(x) = 2x^2 - x - 1$ 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-intercepts, if any, of the graph of $f$.

To find the x-intercepts, we set $f(x) = 0$:
$$2x^2 - x - 1 = 0$$

We can solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -1$, and $c = -1$:

$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)}$$
$$x = \frac{1 \pm \sqrt{1 + 8}}{4}$$
$$x = \frac{1 \pm \sqrt{9}}{4}$$
$$x = \frac{1 \pm 3}{4}$$

This gives us two solutions:
$$x = \frac{1 + 3}{4} = 1$$
$$x = \frac{1 - 3}{4} = -\frac{1}{2}$$

So, the x-intercepts are $x = 1$ and $x = -\frac{1}{2}$. The points are $(1, 0)$ and $\left(-\frac{1}{2}, 0\right)$.

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

To find the y-intercept, we set $x = 0$:
$$f(0) = 2(0)^2 - 0 - 1$$
$$f(0) = -1$$

So, the y-intercept is $y = -1$. The point is $(0, -1)$.

In summary:
(a) Yes, $(-1, 2)$ is on the graph.
(b) When $x = -2$, $f(x) = 9$. The point is $(-2, 9)$.
(c) When $f(x) = -1$, $x = 0$ or $x = \frac{1}{2}$. The points are $(0, -1)$ and $\left(\frac{1}{2}, -1\right)$.
(d) The domain of $f$ is $(-\infty, \infty)$.
(e) The x-intercepts are $(1, 0)$ and $\left(-\frac{1}{2}, 0\right)$.
(f) The y-intercept is $(0, -1)$.