Question

Question 2-5 please

Answer 1

Answer:

1. $m=1\frac{1}{14}$

2. undefined

3. $-6$.

4. $2\frac{1}{3}$.

5. $2$.

6.  $\frac{7}{2}$

Step-by-step explanation:

1. The required line passes through $(3,0),(-11,-15)$. The slope of the line joining the points  $(x_1,y_1),(x_2,y_2)$ is given by the formula,

$m=\frac{y_2-y_1}{x_2-x_1}$ or $m=\frac{y_1-y_2}{x_1-x_2}$

Either of them yields the same result.

Substitute $x_1=3,y_1=0,x_2=-11,y_2=-15$ into the first formula to get,

$m=\frac{-15-0}{-11-3}$ $\implies m=\frac{-15}{-14}=\frac{15}{14}$

You can write this as a mixed number to obtain:

$\implies m=1\frac{1}{14}$

2. The given line passes through $(4,-8),(4,13)$.

Observe that the first coordinates are the same for both points. This implies that, the line is vertical.

The slope of all vertical lines are undefined.

3. The average rate of change of the function $y=f(x)$ on the interval $[a,b]\:\:or\:\:a\le x\le b$ is given by:

$ARC=\frac{f(b)-f(a)}{b-a}$.

The given function is $f(x)=x^2-4x$

$f(-2)=(-2)^2-4(-2)=12$

$f(4)=(4)^2-4(4)=0$

$ARC=\frac{f(4)-f(-2)}{4--2}$.

$ARC=\frac{0-12}{6}=-6$.

4. The given function is $f(x)=2^x-1$

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

$f(3)=2^3-1=7$

$ARC=\frac{f(3)-f(0)}{3-0}$.

$ARC=\frac{7-0}{3}=2\frac{1}{3}$.

5. From the graph

$f(3)=7,f(0)=1$

$ARC=\frac{f(3)-f(0)}{3-0}$.

$ARC=\frac{7-1}{3}$.

$ARC=\frac{6}{3}=2$.

6. The given equation is :

$2y-7x=-4$

Add $7x$ to both sides.

$2y=7x-4$

Divide through by 2.

$y=\frac{7}{2}x-2$

Compare this function to $y=mx+b$

The slope is $m=\frac{7}{2}$