First, use the points to find the equation of JK:

Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

\(\displaystyle m=\frac{y'-y}{x'-x}\)

Plug in and calculate:

\(\displaystyle \small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}\)

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

\(\displaystyle \small 5=\frac{1}{2}*4+b\)

\(\displaystyle \small b=5-2=3\)

So our answer is: 

\(\displaystyle \small y=\frac{x}{2}+3\)

To find y, we need to plug in 16 for x and solve:

\(\displaystyle \small \small y=\frac{16}{2}+3=8+3=11\)

To find out if (-2,5) is on f(x), simply plug the point into f(x)

\(\displaystyle \small f(x)=4x+13\)

Becomes,

\(\displaystyle \small \small 5=4*-2+13=-8+13=5\)

So yes, it does!

One way to solve this one is by plugging in each of the answer choices and eliminating any that don't work out. Begin with our original g(x)

\(\displaystyle {}g(x)=4x^2-8\)

If we plug in 3 we get

\(\displaystyle {}g(x)=4(3)^2-8=4(9)-8=36-8=28\), 

So our point is (3,28).

Since the slope and a point on the line are given, we can use the point-slope formula:

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

\(\displaystyle m=1\ and\ (1,4)\)

\(\displaystyle y-4=1(x-1)\)

\(\displaystyle y-4=x-1\)

\(\displaystyle y=x-1+4\)

\(\displaystyle y=x+3\)

Since the slope and a point on the line are given, we can use the point-slope formula:

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

\(\displaystyle m=0\ and\ (5,2)\)

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

\(\displaystyle y-2=0\)

\(\displaystyle y=2\)

First find the slope of the equation.

\(\displaystyle m =\frac{rise}{run} =\frac{7 + 2}{1-4} = \frac{9}{-3}=-3\)

Now plug in one of the two points to form an equation.  Here we use (4, -2), but either point will produce the same answer.

\(\displaystyle y-(-2)=-3(x-4)\)

\(\displaystyle y + 2=-3x + 12\)

\(\displaystyle y=-3x+10\)

Given that JK passes through (4,5) and (144,75) we can find the slope as follows:

Slope is found via:

\(\displaystyle m=\frac{y'-y}{x'-x}\)

Plug in and calculate:

\(\displaystyle \small \small m=\frac{75-5}{144-4}=\frac{70}{140}=\frac{1}{2}\)

Next, we need to use one of our points and the slope to find our y-intercept. I'll use (4,5).

\(\displaystyle \small 5=\frac{1}{2}*4+b\)

\(\displaystyle \small b=5-2=3\)

So our answer is: 

\(\displaystyle \small y=\frac{x}{2}+3\)

The equation for a line in standard form is written as follows:

\(\displaystyle y=mx+b\)

Where \(\displaystyle m\) is the slope and \(\displaystyle b\) is the y intercept. We start by calculating the slope between the two given points using the following formula:

\(\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{7-3}{6-(-2)}=\frac{4}{8}=\frac{1}{2}\)

Now we can plug either of the given points into the formula for a line with the calculated slope and solve for the y intercept:

\(\displaystyle y=mx+b\)

\(\displaystyle 3=(\frac{1}{2})(-2)+b\)

\(\displaystyle b=4\)

We now have the slope and the y intercept of the line, which is all we need to write its equation in standard form:

\(\displaystyle y=\frac{1}{2}x+4\)

The \(\displaystyle y\)-intercept of the parabola of the equation can be found by substituting 0 for \(\displaystyle x\):

\(\displaystyle y = x^{2} + 5x - 6\)

\(\displaystyle y = 0^{2} + 5 (0) - 6\)

\(\displaystyle y = -6\)

This point is \(\displaystyle (0, -6)\).

The vertex of the parabola of the equation \(\displaystyle y = a x^{2} + bx+c\) has \(\displaystyle x\)-coordinate \(\displaystyle - \frac{b}{2a}\), and its \(\displaystyle y\)-coordinate can be found using substitution for \(\displaystyle x\). Setting \(\displaystyle a = 1\) and \(\displaystyle b= 5\):

\(\displaystyle x = - \frac{b}{2a} = - \frac{5}{2 \cdot 1} = - \frac{5}{2 }\)

\(\displaystyle y = \left ( - \frac{5}{2 } \right )^{2} + 5 \left ( - \frac{5}{2 } \right ) - 6\)

\(\displaystyle y = \frac{25}{4 } - \frac{25}{2 } - 6\)

\(\displaystyle y = \frac{25}{4 } - \frac{50}{4 } - \frac{24}{4}\)

\(\displaystyle y = - \frac{49}{4 }\)

The vertex is \(\displaystyle \left (- \frac{5}{2 },- \frac{49}{4 } \right )\)

The line connects the points \(\displaystyle (0, -6)\) and \(\displaystyle \left (- \frac{5}{2 },- \frac{49}{4 } \right )\). Its slope is

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

\(\displaystyle = \frac{-6-\left ( - \frac{49}{4 } \right )}{0 - \left (- \frac{5}{2 } \right )}\)

\(\displaystyle = \frac{-\frac{24 }{4 } + \frac{49}{4 } }{ \frac{5}{2 } }\)

\(\displaystyle = \frac{ \frac{25}{4 } }{ \frac{5}{2 } }\)

\(\displaystyle = \frac{25}{4 } \cdot \frac{2 } {5}\)

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

Since the line has \(\displaystyle y\)-intercept \(\displaystyle (0, -6)\) and slope \(\displaystyle m =\frac{5}{2 }\), the equation of the line is \(\displaystyle y=\frac{5}{2 } x + (- 6)\), or \(\displaystyle y=\frac{5}{2 } x - 6\).