Q. 23 - Precalculus Enhanced with Graphing Utilities

Question

$f\left(x\right)=2x^2-x-1$

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

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

(c) If $f\left(x\right)=-1$, 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 $f$. 

Step-by-Step Solution

Answer

Part (a) Yes

Part (b) $f\left(-2\right)=9$; The point on the graph is $\left(-2, 9\right)$.

Part (c) Values of $x$ are $0 and \frac{1}{2}$ ; Points on the graph are $\left(0,-1\right) and \left(\frac{1}{2},-1\right)$.

Part (d) The domain is all real numbers.

Part (e) The x-intercepts are $-\frac{1}{2} and 1$.

Part (f) The y-intercept is $-1$.

Part (a) Step 1. Find the value of the function given point.

When $x=-1$,

$\begin{array}{rcl}f\left(x\right)& =& 2x^2-x-1\\ f\left(-1\right)& =& 2{\left(-1\right)}^2-\left(-1\right)-1\\ & =& 2\left(1\right)+1-1\\ & =& 2\end{array}$

The point $\left(-1, 2\right)$ is on the graph of $f$.

Part (b) Step 1. Find the value at -2.

If $x=-2$, then

$\begin{array}{rcl}f\left(x\right)& =& 2x^2-x-1\\ f\left(-2\right)& =& 2{\left(-2\right)}^2-\left(-2\right)-1\\ & =& 2\left(4\right)+2-1\\ & =& 8+1\\ & =& 9\end{array}$

The point $\left(-2, 9\right)$ is on the graph.

Part (c) Step 1. Find the value of x .

If $f\left(x\right)=-1$, then,

$\begin{array}{rcl}f\left(x\right)& =& -1\\ 2x^2-x-1& =& -1\\ 2x^2-x& =& 0\\ x\left(2x-1\right)& =& 0\\ x& =& 0 or x=\frac{1}{2}\end{array}$

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

Part (d) Step 1. Find the domain.

The function is two times square a number and then subtract the number and one. Since these operations can be performed on any real number, the domain of $f$ is the set of all real numbers.

Part (e) Step 1. Find x -intercepts.

The x-intercepts of the graph of $f$ are the real solutions of the equation $f\left(x\right)=0$ that are in the domain of $f$.

The real solution of the equation $f\left(x\right)=2x^2-x-1$is,

$\begin{array}{rcl}f\left(x\right)& =& 0\\ 2x^2-x-1& =& 0\\ \left(x-1\right)\left(2x+1\right)& =& 0\\ x-1& =& 0 or 2x+1=0\\ x& =& 1 or x=-\frac{1}{2}\end{array}$

 The x-intercepts are $1 and -\frac{1}{2}$.

Part (f) Step 1. Find y -intercepts.

The y-intercepts of the graph of $f$ are the real values of $f\left(0\right)$.

$\begin{array}{rcl}f\left(x\right)& =& 2x^2-x-1\\ f\left(0\right)& =& 2{\left(0\right)}^2-\left(0\right)-1\\ & =& 0-1\\ & =& -1\end{array}$

The y-intercept is $-1$.