We are given the quadratic function $f(x) = 2x^2 - x - 1$, and we'll answer each part of the question accordingly.

(a) Is the point $(-2, 9)$ on the graph of $f$?

To check if the point $(-2, 9)$ is on the graph of the function, substitute $x = -2$ into $f(x)$ and see if it equals 9.

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

Since $f(-2) = 9$, the point $(-2, 9)$ is on the graph of $f$.

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

Substitute $x = 2$ into the function to find $f(2)$:

$f(2) = 2(2)^2 - 2 - 1 = 2(4) - 2 - 1 = 8 - 2 - 1 = 5$

Thus, $f(2) = 5$, and the point $(2, 5)$ is on the graph of $f$.

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

To find $x$ when $f(x) = -1$, set $f(x) = -1$ and solve for $x$:

$2x^2 - x - 1 = -1$

Simplify the equation:

$2x^2 - x - 1 + 1 = 0 \quad \Rightarrow \quad 2x^2 - x = 0$

Factor the equation:

$x(2x - 1) = 0$

The solutions are:

$x = 0 \quad \text{or} \quad 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2}$

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

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

Since $f(x)$ is a quadratic function, it is defined for all real values of $x$. Therefore, the domain of $f$ is:

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

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

The $x$-intercepts occur where $f(x) = 0$. Set $f(x) = 0$ and solve:

$2x^2 - x - 1 = 0$

Solve this quadratic equation using the quadratic formula:

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

So,

$x = \frac{1 + 3}{4} = 1 \quad \text{or} \quad x = \frac{1 - 3}{4} = -\frac{1}{2}$

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

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(f) List the $y$-intercept, if any, of the graph of $f$.

The $y$-intercept occurs where $x = 0$. Substituting $x = 0$ into $f(x)$:

$f(0) = 2(0)^2 - 0 - 1 = -1$

Thus, the $y$-intercept is $(0, -1)$.

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Summary of Answers:

• (a) Yes, the point $(-2, 9)$ is on the graph.
• (b) $f(2) = 5$; the point is $(2, 5)$.
• (c) $x = 0$ 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 $(0, -1)$.

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Do you need any further details on any of the parts?

Related questions:

1. What is the vertex of the quadratic function $f(x)$?
2. What is the axis of symmetry for the function $f(x)$?
3. How do you find the maximum or minimum value of a quadratic function?
4. What is the range of $f(x) = 2x^2 - x - 1$?
5. How do you graph the function $f(x) = 2x^2 - x - 1$?

Tip:

To solve quadratic equations efficiently, the quadratic formula is a powerful tool for finding roots when factoring is difficult.