Chapter 2.2, Problem 24AYU

In Problems 25-30, answer the questions about the given function.

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

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

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

(c) If $f\left(x\right)\text{ = }-2$, what is $x$? What point(s) are on the graph of $f$?

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

(e) List the $x\text{-intercepts}$, if any, of the graph of $f$.

(f) List the $y\text{-intercepts}$, if there is one, of the graph of $f$.

To determine

To find: When $f\left(x\right)=-3x²+5x$, answer the following

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 of $x\text{-intercepts}$, if any, of the graph of $f$.

f. List of the $y\text{-intercept}$, if there is one, of the graph of $f$.

Answer to Problem 24AYU

Solution:

a. No, the point $\left(-1,2\right)$ is not on the graph of $f$.

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

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

d. Domain of $f$ = All real numbers.

e. $x=0, x=\frac{5}{3}$.

f. $y\text{-intercept}=0$.

Explanation of Solution

Given:

$f\left(x\right)=-3x²+5x$

Calculation:

a. $f\left(x\right)=-3x²+5x$

To find the point $\left(-1,2\right)$ on the graph of $f$, substitute $x=-1$ in $f(x)$.

$f\left(-1\right)=-3\left(-1\right)²+5\left(-1\right)$

$=-3\times 1-5$

$=-3-5$

$=-8\ne 2$

No, the point $\left(-1,2\right)$ is not on the graph of $f$.

b. If $x=-2$.

$f(-2)=-3\left(-2\right)²+5(-2)$

$=-3\times 4-10$

$=-12-10$

$=-22$

The point $(-2,-22)$ is on the graph of $f$.

c. $f\left(x\right)=-2\Rightarrow -3x²+5x =-2$

$-3x²+5x+2=0$

$3x²-5x-2=0$

$3x²-6x+x-2=0$

$3x\left(x-2\right)+\left(x-2\right)=0$

$\left(3x+1\right)\left(x-2\right)=0$

$x-2=0,3x+1=0$

$x=2,x=-\frac{1}{3}$

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

d. Domain of $f$ = All real numbers.

e. To find $x\text{-intercept}$, let $f\left(x\right)=0$,

$-3x²+5x=0$

$x\left(5-3x\right)=0$

$x=0,5-3x=0$

$x=0,x=\frac{5}{3}$

f. To find the $y\text{-intercept}$, let $x=0$ in $f\left(x\right)=-3x²+5x$.

$f\left(0\right)=-3\left(0\right)²+5\left(0\right) =0\Rightarrow y\text{-intercept}=0$