To find the slope, you will need to put the equation in $\displaystyle y=mx+b$ form. The value of $\displaystyle m$ will be the slope.

$\displaystyle 12x-8y=6$

Subtract $\displaystyle 6$ from either side:

$\displaystyle 8y=12x-6$

Divide each side by $\displaystyle 8$:

$\displaystyle y=\frac{3}{2}x-\frac{3}{4}$

You can now easily identify the value of $\displaystyle m$.

$\displaystyle m=\frac{3}{2}$

The $\displaystyle y$-intercept of the graph of a function is the point at which it intersects the $\displaystyle y$-axis - that is, at which $\displaystyle x = 0$. This point is $\displaystyle \left (0, f(0) \right )$, so evaluate $\displaystyle f(0)$:

$\displaystyle f \left ( x\right ) = 2x^{2} - 7x + 5$

$\displaystyle f \left (0\right ) = 2 \cdot 0^{2} - 7 \cdot 0 + 5$

$\displaystyle f \left (0\right ) = 0- 0 + 5 = 5$

The $\displaystyle y$-intercept is $\displaystyle \left (0,5 \right )$.

The $\displaystyle y$-intercept is the point at which the graph crosses the $\displaystyle y$-axis; at this point, the $\displaystyle x$-coordinate is 0, so substitute $\displaystyle 0$ for $\displaystyle x$ in the equation:

$\displaystyle y = \frac{1}{x^{2}+ x} + 3$

$\displaystyle y = \frac{1}{0^{2}+0} + 3$

$\displaystyle y = \frac{1}{0} + 3$

However, since this expression has 0 in a denominator, it is of undefined value. This means that there is no value of $\displaystyle x$ paired with $\displaystyle y$-coordinate 0, and, subsequently, the graph of the equation has no $\displaystyle y$-intercept.

The $\displaystyle y$-intercept is the point at which the graph crosses the $\displaystyle y$-axis; at this point, the $\displaystyle x$-coordinate is 0, so substitute $\displaystyle 0$ for $\displaystyle x$ in the equation:

$\displaystyle y = \frac{3}{\left | 3x-8\right |}$

$\displaystyle y = \frac{3}{\left | 3 \cdot 0 -8\right |}$

$\displaystyle y = \frac{3}{\left | 0 -8\right |}$

$\displaystyle y = \frac{3}{\left | -8\right |}$

$\displaystyle y = \frac{3}{ 8}$

The $\displaystyle y$-intercept is $\displaystyle \left ( 0, \frac{3}{8}\right )$.

The $\displaystyle y$-intercept is the point at which the graph crosses the $\displaystyle y$-axis; at this point, the $\displaystyle x$-coordinate is 0, so substitute $\displaystyle 0$ for $\displaystyle x$ in the equation:

$\displaystyle y = 3(x-2)^{2} + 4 (x-7)$

$\displaystyle y = 3(0-2)^{2} + 4 (0-7)$

$\displaystyle y = 3( -2)^{2} + 4 ( -7)$

$\displaystyle y = 3 \cdot 4 + 4 ( -7)$

$\displaystyle y = 12+ 4 ( -7)$

$\displaystyle y = 12+ (-28)$

$\displaystyle y = -16$

The $\displaystyle y$-intercept is the point $\displaystyle (0, -16)$. 

First, find the slope of the second line $\displaystyle 3x-4y = 17$ by solving for $\displaystyle y$ as follows:

$\displaystyle 3x-4y = 17$

$\displaystyle 3x-4y +4y - 17 = 17+4y - 17$

$\displaystyle 4y = 3x-17$

$\displaystyle \frac{4y }{4}=\frac{ 3x-17 }{4}$

$\displaystyle y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form $\displaystyle y = mx+ b$; the slope of the second line is the $\displaystyle x$-coefficient $\displaystyle m= \frac{3}{4}$.

The first line, being perpendicular to the second, has as its slope the opposite of the reciprocal of $\displaystyle \frac{3}{4}$, which is $\displaystyle m=- \frac{4}{3}$.

Therefore, we are looking for a line through $\displaystyle (5, 2)$ with slope $\displaystyle m=- \frac{4}{3}$. Using point-slope form

$\displaystyle y - y_{1} = m(x -x_{1})$

with 

$\displaystyle x _{1 } =5, y _{1 } = 2,m=- \frac{4}{3}$,

the equation becomes

$\displaystyle y - 2 =- \frac{4}{3}(x -5)$.

To find the $\displaystyle x$-intercept, substitute 0 for $\displaystyle y$ and solve for $\displaystyle x$:

$\displaystyle 0 - 2 =- \frac{4}{3}(x -5)$

$\displaystyle - 2 =- \frac{4}{3}x+ \frac{20}{3}$

$\displaystyle - 2- \frac{20}{3} =- \frac{4}{3}x+ \frac{20}{3} - \frac{20}{3}$

$\displaystyle - \frac{26}{3} =- \frac{4}{3}x$

$\displaystyle - \frac{3} {4}\cdot\left (- \frac{26}{3} \right )=- \frac{3} {4}\cdot \left (- \frac{4}{3}x \right )$

$\displaystyle x= \frac{26}{4} = \frac{13}{2} = 6\frac{1}{2}$

The  $\displaystyle x$-intercept is the point $\displaystyle \left (6 \frac{1}{2}, 0 \right )$.

First, find the slope of the second line $\displaystyle 3x-4y = 17$ by solving for $\displaystyle y$ as follows:

$\displaystyle 3x-4y = 17$

$\displaystyle 3x-4y +4y - 17 = 17+4y - 17$

$\displaystyle 4y = 3x-17$

$\displaystyle \frac{4y }{4}=\frac{ 3x-17 }{4}$

$\displaystyle y = \frac{3}{4}x- \frac{17}{4}$

The equation is now in the slope-intercept form $\displaystyle y = mx+ b$; the slope of the second line is the $\displaystyle x$-coefficient $\displaystyle m= \frac{3}{4}$.

The first line, being parallel to the second, has the same slope. 

Therefore, we are looking for a line through $\displaystyle (5, 2)$ with slope $\displaystyle m= \frac{3}{4}$. Using point-slope form

$\displaystyle y - y_{1} = m(x -x_{1})$

with 

$\displaystyle x _{1 } =5, y _{1 } = 2,m= \frac{3}{4}$,

the equation becomes

$\displaystyle y - 2 = \frac{3}{4}(x -5)$.

To find the $\displaystyle x$-intercept, substitute 0 for $\displaystyle y$ and solve for $\displaystyle x$:

$\displaystyle 0 - 2 = \frac{3}{4}(x -5)$

$\displaystyle - 2 = \frac{3}{4} x - \frac{15}{4}$

$\displaystyle - 2 + \frac{15}{4}= \frac{3}{4} x - \frac{15}{4} + \frac{15}{4}$

$\displaystyle \frac{3}{4} x = \frac{7}{4}$

$\displaystyle \frac{4} {3}\cdot \frac{3}{4} x =\frac{4} {3}\cdot \frac{7}{4}$

$\displaystyle x = \frac{7}{3} = 2\frac{1}{3}$

The $\displaystyle x$-intercept is the point $\displaystyle \left (2 \frac{1}{3}, 0 \right )$.

By the point-slope formula, this line has the equation

$\displaystyle y - y_{1} = m(x -x_{1})$

where

$\displaystyle x_{1} = 5, y_{1} = 9 ,m = \frac{2}{5}$

By substitution, the equation becomes

$\displaystyle y -9 = \frac{2}{5}(x -5)$ 

To find the $\displaystyle y$-intercept, substitute 0 for $\displaystyle x$ and solve for $\displaystyle y$:

$\displaystyle y -9 = \frac{2}{5}(0 -5)$

$\displaystyle y -9 = \frac{2}{5}( -5)$

$\displaystyle y - 9 = -2$

$\displaystyle y = 7$

The $\displaystyle y$-intercept is the point $\displaystyle (0, 7)$.

First, find the slope of the line, using the slope formula

$\displaystyle m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$

setting $\displaystyle x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$:

$\displaystyle m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

$\displaystyle y - y_{1} = m(x -x_{1})$

where

$\displaystyle x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$; the line becomes

$\displaystyle y - 4 = - \frac{7}{5}(x -(-3))$

or

$\displaystyle y - 4 = - \frac{7}{5}(x +3)$

To find the $\displaystyle y$-intercept, substitute 0 for $\displaystyle x$ and solve for $\displaystyle y$:

$\displaystyle y - 4 = - \frac{7}{5}(0+3)$

$\displaystyle y - 4 = - \frac{7}{5}(3)$

$\displaystyle y - 4 = - \frac{21}{5}$

$\displaystyle y - 4+ 4 = - \frac{21}{5} + 4$

$\displaystyle y = - \frac{1}{5}$

The  $\displaystyle y$-intercept is $\displaystyle \left (0, - \frac{1}{5} \right )$.

First, find the slope of the line, using the slope formula

$\displaystyle m = \frac{y_{2} -y_{1}}{x_{2}-x_{1}}$

setting $\displaystyle x_{1} = -3, y_{1} = 4, x_{2} = 2, y_{2} = -3$:

$\displaystyle m = \frac{-3 -4}{2-(-3)} = \frac{-7}{5} =- \frac{7}{5}$

By the point-slope formula, this line has the equation

$\displaystyle y - y_{1} = m(x -x_{1})$

where

$\displaystyle x_{1} = -3, y_{1} = 4, m = - \frac{7}{5}$; the line becomes

$\displaystyle y - 4 = - \frac{7}{5}(x -(-3))$

or

$\displaystyle y - 4 = - \frac{7}{5}(x +3)$

To find the $\displaystyle x$-intercept, substitute 0 for $\displaystyle y$ and solve for $\displaystyle x$:

$\displaystyle 0 - 4 = - \frac{7}{5}(x +3)$

$\displaystyle - 4 = - \frac{7}{5} x - \frac{21}{5}$

$\displaystyle - 4+ \frac{21}{5} = - \frac{7}{5} x - \frac{21}{5}+ \frac{21}{5}$

$\displaystyle \frac{1}{5} = - \frac{7}{5} x$

$\displaystyle - \frac{5}{7} \cdot \frac{1}{5} = - \frac{5}{7}\left ( - \frac{7}{5} x \right )$

$\displaystyle - \frac{1}{7} =x$

The  $\displaystyle x$-intercept is $\displaystyle \left (- \frac{1}{7} , 0 \right )$.