Q. 24 - Precalculus Enhanced with Graphing Utilities

Question

$f\left(x\right)=-3x^2+5x$

(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 $f\left(x\right)=-2$, 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) No

Part (b) $f\left(-2\right)=-22$; The point on the graph is $\left(-2,-22\right)$.

Part (c) Values of $x$ are $2 and -\frac{1}{3}$; Points on the graph are $\left(2,-2\right) and \left(-\frac{1}{3},-2\right)$.

Part (d) The domain is all real numbers.

Part (e) The x-intercepts are $0 and \frac{5}{3}$.

Part (f) The y-intercept is $0$.

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

When $x=-1$,

$\begin{array}{rcl}f\left(x\right)& =& -3x^2+5x\\ f\left(-1\right)& =& -3{\left(-1\right)}^2+5\left(-1\right)\\ & =& -3\left(1\right)-5\\ & =& -8\end{array}$

The point  is not on the graph of $f$.

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

If $x=-2$,

$\begin{array}{rcl}f\left(x\right)& =& -3x^2+5x\\ f\left(-2\right)& =& -3{\left(-2\right)}^2+5\left(-2\right)\\ & =& -3\left(4\right)-10\\ & =& -12-10\\ & =& -22\end{array}$

The point  is on the graph.

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

If $f\left(x\right)=-2$,

$\begin{array}{rcl}f\left(x\right)& =& -2\\ -3x^2+5x& =& -2\\ 3x^2-5x-2& =& 0\\ \left(x-2\right)\left(3x+1\right)& =& 0\\ x-2& =& 0 or 3x+1=0\\ x& =& 2 or x=-\frac{2}{3}\end{array}$

The point on the graph are $\left(2,-2\right) and \left(-\frac{1}{3},-2\right)$.

Part (d) Step 1. Find the domain.

The function is five times the number and then subtract three times square a number. 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 $\begin{array}{rcl}f\left(x\right)& =& -3x^2+5x\end{array}$ is

1. $\begin{array}{rcl}f\left(x\right)& =& 0\\ -3x^2+5x& =& 0\\ -x\left(3x-5\right)& =& 0\\ x& =& 0 or 3x-5=0\\ x& =& 0 or x=\frac{5}{3}\end{array}$
   

 The x-intercepts are $0 and \frac{5}{3}$.

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

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

$\begin{array}{rcl}f\left(x\right)& =& -3x^2+5x\\ f\left(0\right)& =& -3{\left(0\right)}^2+5\left(0\right)\\ & =& 0+0\\ & =& 0\end{array}$

The y-intercept is $0$.