# What is the slope and y-intercept in case of following cases representing lines?

## $f \left(x\right)$ passes through $\left(- 2 , 2\right)$ and also through $\left(0 , - 4\right)$ $g \left(x\right)$ is given by $y = 3 x - 4$ $h \left(x\right)$ represents apples sold on Monday as $4$ and on Tuesday, it is $1$ $j \left(x\right)$ moves through points $\left(- 2 , - 2\right)$, $\left(2 , 10\right)$, $\left(6 , 221\right)$

Sep 23, 2016

$f \left(x\right)$ - slope is $- 3$ and $y$-intercept is $- 4$
$g \left(x\right)$ - slope is $3$ and $y$-intercept is $- 4$
$h \left(x\right)$ - slope is $- 3$ and $y$-intercept is $4$
$j \left(x\right)$ - slope is $3$ and $y$-intercept is $4$

#### Explanation:

$f \left(x\right)$ - This line passes through $\left(- 2 , 2\right)$ and $y$-intercept is $- 4$ i.e. it passes through $\left(0 , - 4\right)$. Hence its slope is $\frac{- 4 - 2}{0 - \left(- 2\right)} = - \frac{6}{2} = - 3$. Hence $f \left(x\right)$ is given by $y = - 3 x - 4$

$g \left(x\right)$ - As give this is given by $y = 3 x - 4$, hence slope is $3$ and intercept on $y$-axis is $- 4$.

$h \left(x\right)$ - here we assume that Monday is $0$ and Tuesday is $1$, hence as number of apples on $0$ day is $4$ and on $1$ day is $1$, the representative $y$ intercept is $4$ and slope is $\frac{1 - 4}{1 - 0} = - \frac{3}{1} = - 3$.

$j \left(x\right)$ - here as the $x$ in function moves from $- 2$ to $2$ and then $6$ and $j \left(x\right)$ moves from $- 2$ to $10$ and then $22$, the slope is $\frac{10 - \left(- 2\right)}{2 - \left(- 2\right)} = \frac{12}{4} = 3$ also $\frac{22 - 10}{6 - 2} = \frac{12}{4} = 3$. For $y$-intercept, we can take mid point of $2$ and $- 2$, where $j \left(x\right) = \frac{10 + \left(- 2\right)}{2} = \frac{8}{2} = 4$ and $y$-intercept is $4$