Let \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of the line. Then the slope of the line is \(\displaystyle -\frac{b}{a}\), or, equilvalently, \(\displaystyle - (b\div a)\). 

We do not need to find the actual slopes of the four choices if we observe that in each case, \(\displaystyle a\) and \(\displaystyle b\) are of the same sign. Since the quotient of two numbers of the same sign is positive, it follows that \(\displaystyle - (b\div a)\) is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope \(\displaystyle \frac{3}{4}\).

 Let \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of the line. Then the slope of the line is \(\displaystyle -\frac{b}{a}\), or, equilvalently, \(\displaystyle - (b\div a)\). 

We can examine the intercepts in each choice to determine which set meets these conditions.

\(\displaystyle \left ( \frac{5}{6}, 0 \right )\) and \(\displaystyle \left ( 0, \frac{5}{8}\right )\):

Slope: \(\displaystyle - \left (b\div a \right )=- \left (\frac{5}{8}\div \frac{5}{6} \right )= - \frac{3}{4}\)

\(\displaystyle \left ( \frac{6}{7},0 \right )\) and \(\displaystyle \left ( 0, \frac{4}{7} \right )\)

Slope: \(\displaystyle - \left (b\div a \right )=- \left ( \frac{4}{7}\div \frac{6}{7} \right )=-\frac{2}{3}\)

\(\displaystyle \left ( \frac{3}{7},0 \right )\) and \(\displaystyle \left ( 0, \frac{4}{7} \right )\)

Slope: \(\displaystyle - \left (b\div a \right )= - \left ( \frac{4}{7}\div \frac{3}{7} \right )= -\frac{4}{3}\)

\(\displaystyle \left ( \frac{4}{5}, 0 \right )\) and \(\displaystyle \left ( 0, \frac{3}{5}\right )\)

Slope: \(\displaystyle - \left (b\div a \right )=- \left ( \frac{3}{5}\div \frac{4}{5} \right )= -\frac{3}{4}\)

\(\displaystyle \left ( \frac{3}{7},0 \right )\) and \(\displaystyle \left ( 0, \frac{4}{7} \right )\) comprise the correct choice.

The slope of the green line can be calculated by noting that the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts of the line are, respectively, \(\displaystyle (3,0)\) and \(\displaystyle (0,5)\).  If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of a line, the slope of the line is \(\displaystyle -\frac{b}{a}\). This makes the slope of the green line \(\displaystyle -\frac{5}{3}\). 

Any line perpendicular to this line must have as its slope the opposite reciprocal of this, or \(\displaystyle \frac{3}{5}\). Since the desired line must also have \(\displaystyle x\)-intercept \(\displaystyle (3,0)\), the equation of the line, in point=slope form, is

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

\(\displaystyle y -0 = \frac{3}{5} (x-3)\)

which can be simplified as

\(\displaystyle y = \frac{3}{5} x-\frac{9}{5}\)

\(\displaystyle 5 \cdot y =5 \cdot \left ( \frac{3}{5} x-\frac{9}{5} \right )\)

\(\displaystyle 5 y =3 x-9\)

\(\displaystyle 5 y- 5y + 9 =3 x-9 - 5y + 9\)

\(\displaystyle 3 x - 5y = 9\)

To locate the \(\displaystyle y \:\)-intercept of the equation \(\displaystyle y = x^{2} - 8x+ 15\), substitute 0 for \(\displaystyle x\):

\(\displaystyle y = x^{2} - 8x+ 15\)

\(\displaystyle y = 0^{2} - 8 \cdot 0+ 15\)

\(\displaystyle y = 15\)

The \(\displaystyle y \:\)-intercept of the parabola is \(\displaystyle (0,15)\).

The vertex of the parabola of an equation of the form \(\displaystyle y = ax^{2}+ bx + c\) has \(\displaystyle x\)-coordinate \(\displaystyle -\frac{b}{2a}\). Here, we substitute \(\displaystyle a = 1, b = -8\), to obtain \(\displaystyle x\)-coordinate

\(\displaystyle -\frac{b}{2a} = -\frac{-8}{2 (1)} = 4\).

To find the \(\displaystyle y \:\)-coordinate, substitute this for \(\displaystyle x\):

\(\displaystyle y = x^{2} - 8x+ 15\)

\(\displaystyle y = 4^{2} - 8 \cdot 4+ 15 = 16 - 32 +15 = -1\)

The vertex is \(\displaystyle (4,-1)\).

The line includes points \(\displaystyle (0,15)\) and \(\displaystyle (4,-1)\); apply the slope formula:

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

\(\displaystyle m = \frac{15-(-1)}{0-4} = \frac{16}{-4} = -4\)

The slope is \(\displaystyle -4\), and the \(\displaystyle y \;\)-intercept is \(\displaystyle (0,15)\); in the slope-intercept form \(\displaystyle y = mx+b\), substitute for \(\displaystyle m = -4\) and \(\displaystyle b = 15\). The equation of the line is \(\displaystyle y = -4x+15\).

A line with undefined slope is a vertical line, and its equation is \(\displaystyle x = a\) for some \(\displaystyle a\), so the \(\displaystyle x\)-coordinate of all points it passes through is \(\displaystyle a\). If it goes through the vertex of a parabola \(\displaystyle (h,k )\), then the line has the equation \(\displaystyle x=h\). Therefore, all we need to find is the \(\displaystyle x\)-coordinate of the vertex of the parabola.

The vertex of the parabola of the equation \(\displaystyle y = ax^{2}+ bx+c\) has as its \(\displaystyle x\)-coordinate \(\displaystyle h=-\frac{b}{2a}\), which, for the parabola of the equation \(\displaystyle y = - x^{2}- 14 x +16\), can be found by setting \(\displaystyle a= -1, b=-14\):

\(\displaystyle h = -\frac{b}{2a} = -\frac{-14}{2(-1)} = -7\)

The desired line is \(\displaystyle x =- 7\).

 Let \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of the line. Then the slope of the line is \(\displaystyle -\frac{b}{a}\), or, equilvalently, \(\displaystyle - (b\div a)\). 

We can examine the intercepts in each choice to determine which set meets these conditions.

\(\displaystyle (12.8 , 0)\) and \(\displaystyle (0, 3.2)\)

Slope:  \(\displaystyle - \left (b\div a \right )=- \left (3.2\div 12.8 \right )=-0.25\)

\(\displaystyle (4.3, 0)\) and \(\displaystyle (0, 17.2)\)

Slope:  \(\displaystyle - \left (b\div a \right )=- \left (17.2\div 4.3 \right )=-4\)

\(\displaystyle (12.4, 0)\) and \(\displaystyle (0, -3.1)\)

Slope:  \(\displaystyle - \left (b\div a \right )= - \left (-3.1 \div 12.4 \right )= 0.25\)

\(\displaystyle (2.7, 0)\) and \(\displaystyle (0, -10.8)\)

Slope:  \(\displaystyle - \left (b\div a \right )=- \left (-10.8\div 2.7 \right )=4\)

\(\displaystyle (2.7, 0)\) and \(\displaystyle (0, -10.8)\) comprise the correct choice, since a line passing through these points has the correct slope.

The \(\displaystyle y \,\)-intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the \(\displaystyle y \,\)-intercept of the parabola, substitute 0 for \(\displaystyle x\) in the quadratic equation:

\(\displaystyle y =-6x^{2}+ 11x + 10\)

\(\displaystyle y =-6 \cdot 0^{2}+ 11 \cdot 0+ 10\)

\(\displaystyle y = 10\)

The \(\displaystyle y \,\)-intercept of the parabola, and of the line, is \(\displaystyle (0,10)\).

The \(\displaystyle x\)-intercept of the line coincides with one of those of the parabola; to find the \(\displaystyle x\)-intercepts of the parabola, substitute 0 for \(\displaystyle y \,\) in the equation:

\(\displaystyle y =-6x^{2}+ 11x + 10\)

\(\displaystyle -6x^{2}+ 11x + 10= 0\)

\(\displaystyle -\left (-6x^{2}+ 11x + 10 \right )=- 0\)

\(\displaystyle 6x^{2}- 11x -10= 0\)

Using the \(\displaystyle ac\) method, split the middle term by finding two integers whose product is \(\displaystyle 6 (-10)= -60\) and whose sum is \(\displaystyle -11\); by trial and error we find these to be \(\displaystyle -15\) and 4, so proceed as follows:

\(\displaystyle 6x^{2}-15x+4 x -10= 0\)

\(\displaystyle (6x^{2}-15x)+(4 x -10)= 0\)

\(\displaystyle 3x(2x-5)+2(2x-5)= 0\)

\(\displaystyle (3x +2)(2x-5)= 0\)

Split:

\(\displaystyle 3x+2 = 0\)

\(\displaystyle 3x= -2\)

\(\displaystyle x= -\frac{2}{3}\)

or

\(\displaystyle 2x-5 = 0\)

\(\displaystyle 2x= 5\)

\(\displaystyle x = \frac{5}{2}\)

The \(\displaystyle x\)-intercepts of the parabola are \(\displaystyle \left ( -\frac{2}{3}, 0 \right )\) and \(\displaystyle \left ( \frac{5}{2}, 0 \right )\), so the \(\displaystyle x\)-intercept of the line is one of these. We examine both possibilities.

If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of the line, then the slope of the line is \(\displaystyle -\frac{b}{a}\), or, equivalently, \(\displaystyle - \left ( b \div a\right )\)

If the intercepts are \(\displaystyle \left ( -\frac{2}{3}, 0 \right )\) and \(\displaystyle (0,10)\), the slope is \(\displaystyle - \left [ 10 \div \left (- \frac{2}{3} \right )\right ] = 15\); if the intercepts are \(\displaystyle \left ( \frac{5}{2}, 0 \right )\) and \(\displaystyle (0,10)\), the slope is \(\displaystyle - \left ( 10 \div \frac{5}{2}\right ) = -4\). Since the line is of negative slope, we choose the line of slope \(\displaystyle -4\); since its \(\displaystyle y \,\)-intercept is \(\displaystyle (0,10)\), then we can substitute \(\displaystyle m = -4, b = 10\) in the slope-intercept form of the line, \(\displaystyle y = mx+b\), to get the correct equation, \(\displaystyle y = -4x+10\).

If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of a line, the slope of the line is \(\displaystyle -\frac{b}{a}\). 

The \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts of the line are, respectively, \(\displaystyle (3,0)\) and \(\displaystyle (0,5)\), so \(\displaystyle a =3, b=5\), and consequently, the slope of the green line is \(\displaystyle -\frac{5}{3}\).  A line parallel to this line must also have slope \(\displaystyle m = -\frac{5}{3}\). 

Each of the equations of the lines is in slope-intercept form \(\displaystyle y = mx+b\), where \(\displaystyle m\) is the slope, so we need only look at the coefficients of \(\displaystyle x\). The only choice that has \(\displaystyle -\frac{5}{3}\) as its \(\displaystyle x\)-coefficient is \(\displaystyle y = -\frac{5}{3}x - 7\), so this is the correct choice.

The \(\displaystyle y \,\)-intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the \(\displaystyle y \,\)-intercept of the parabola, substitute 0 for \(\displaystyle x\) in the quadratic equation:

\(\displaystyle y = x^{2}+7x - 18\)

\(\displaystyle y = 0^{2}+7 \cdot 0 - 18\)

\(\displaystyle y = - 18\)

The \(\displaystyle y \,\)-intercept of the parabola, and of the line, is \(\displaystyle (0,-18)\).

The \(\displaystyle x\)-intercept of the line coincides with one of those of the parabola; to find the \(\displaystyle x\)-intercepts of the parabola, substitute 0 for \(\displaystyle y \,\) in the equation:

\(\displaystyle y = x^{2}+7x - 18\)

\(\displaystyle x^{2}+7x - 18 = 0\)

The quadratic expression can be "reverse-FOILed" by noting that 9 and \(\displaystyle -2\) have product \(\displaystyle -18\) and sum 7:

\(\displaystyle (x+9)(x-2)= 0\)

\(\displaystyle x+9 = 0\), in which case \(\displaystyle x= -9\) 

or

\(\displaystyle x-2 = 0\), in which case \(\displaystyle x = 2\).

The \(\displaystyle x\)-intercepts of the parabola are \(\displaystyle (-9,0)\) and \(\displaystyle (2,0)\), so the \(\displaystyle x\)-intercept of the line is one of these. We will examine both possibilities

If \(\displaystyle (a,0)\) and \(\displaystyle (0,b)\) be the \(\displaystyle x\)- and \(\displaystyle y\,\)-intercepts, respectively, of the line, then the slope of the line is \(\displaystyle -\frac{b}{a}\). If the intercepts are \(\displaystyle (2,0)\) and \(\displaystyle (0,-18)\), the slope is \(\displaystyle -\frac{-18}{2}= 9\); if the intercepts are \(\displaystyle (-9,0)\) and \(\displaystyle (0,-18)\), the slope is \(\displaystyle -\frac{-18}{-9}= -2\). Since the line is of positive slope, we choose the line of slope 9; since its \(\displaystyle y \,\)-intercept is \(\displaystyle (0,-18)\), then we can substitute \(\displaystyle m = 9, b = -18\) in the slope-intercept form of the line, \(\displaystyle y = mx+b\), to get the correct equation, \(\displaystyle y = 9x - 18\).

We can find out whether the graphs of \(\displaystyle x+y = 2\) and \(\displaystyle y = 2x^{2} - 7x+5\) intersect by first solving for \(\displaystyle y \;\) in the first equation:

\(\displaystyle x+y = 2\)

\(\displaystyle y = 2 -x\)

We then substitute in the second equation for \(\displaystyle y \;\):

\(\displaystyle 2x^{2} - 7x+5 = 2-x\)

Then we rewrite in standard form:

\(\displaystyle 2x^{2} - 7x+5 -2 + x = 2-x -2 + x\)

\(\displaystyle 2x^{2} - 6x+3 = 0\)

Since we are only trying to pdetermine whether at least one point of intersection exists, rather than actually find the point, all we need to do is to evaluate the discriminant; if it is nonnegative, at least one solution - and, consequently, one point of intersection - exists. In the general quadratic equation \(\displaystyle ax^{2} +bx+c = 0\), this is \(\displaystyle b^{2} - 4ac\), so here, the discriminant is

\(\displaystyle b^{2} - 4ac = (-6)^{2}- 4(2)(3) = 36-24 = 12\ge 0\).

Therefore, the line of the equation \(\displaystyle x+y = 2\) intersects the parabola of the equation \(\displaystyle y = 2x^{2} - 7x+5\).

We do the same for the other three lines:

\(\displaystyle x+y = 4\)

\(\displaystyle y = 4-x\)

\(\displaystyle 2x^{2} - 7x+5 = 4-x\)

Then we rewrite in standard form:

\(\displaystyle 2x^{2} - 7x+5 -4+ x = 4-x -4 + x\)

\(\displaystyle 2x^{2} - 6x+1 = 0\)

\(\displaystyle b^{2} - 4ac = (-6)^{2}- 4(2)(1) = 36-8= 28\ge 0\).

The line of \(\displaystyle x+y = 4\) intersects the parabola.

\(\displaystyle x-y = 2\)

\(\displaystyle x = y + 2\)

\(\displaystyle y = x-2\)

\(\displaystyle 2x^{2} - 7x+5 = x-2\)

\(\displaystyle 2x^{2} - 7x+5- x+2 = x-2 - x+2\)

\(\displaystyle 2x^{2} - 8x+7 =0\)

\(\displaystyle b^{2} - 4ac = (-8)^{2}- 4(2)(7) = 64-56= 8\ge 0\)

The line of \(\displaystyle x-y = 2\) intersects the parabola.

\(\displaystyle x-y = 4\)

\(\displaystyle x = y + 4\)

\(\displaystyle y = x-4\)

\(\displaystyle 2x^{2} - 7x+5 = x-4\)

\(\displaystyle 2x^{2} - 7x+5- x+4 = x-4 - x+4\)

\(\displaystyle 2x^{2} - 8x+9 =0\)

\(\displaystyle b^{2} - 4ac = (-8)^{2}- 4(2)(9) = 64-72= -8 < 0\)

Since the discriminant is negative, the system has no real solution. This means that the line of \(\displaystyle x-y = 4\) does not intersect the parabola of the equation \(\displaystyle y = 2x^{2} - 7x+5\), and it is the correct choice.