Question

Part A: A highway's path can be found using the equation 2x - 3y = 21. Use the graphs of the functions to determine the number of intersections there will be between the railroad and the highway, and explain completely.

Part B: A turnpike's route is determined by the equation y=1/3x^2

Prove algebraically how many intersections there will be between the railroad and the turnpike, showing all necessary work.

Answer 1

We want to find the intersections of different equations in two different ways, graphically and algebraically.

The solutions are:

• a) There are no intersections.
• b) There are two intersections, at (5, 25/3) and (-3, 3).

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Let's see how to get the solutions.

First, remember that a general linear equation in the slope-intercept form is written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

If we know that the line passes through two points (x₁, y₁) and (x₂, y₂) then the slope can be written as:

$a = \frac{y_2 - y_1}{x_2 - x_1}$

We can look at the given graph and see that the line passes through the points (0, 5) and (-3, 3)

Then the slope of this line will be:

$a = \frac{5 - 3}{0 - (-3)} = \frac{2}{3}$

Then this line is something like:

y = (2/3)*x + b

To find the value of b, we use the fact that the line passes through the point (0, 5). This means that when x = 0, we have y = 5.

Then we have:

5 = (2/3)*0 + b

5 = b

Thus the equation of the railroad is:

y = (2/3)*x + 5

a) We want to find the intersection of the graphed line with 2x - 3y = 21, here you only need to graph both lines in the same coordinate axis.

To graph a line, you can write it in slope-intercept form as:

-2x + 3y = -21

3y = 2x - 21

y = (2/3)*x - 21/3

y = (2/3)*x - 7

Notice that this line has the same slope as the other, so these lines are parallel, thus there is no intercept between these lines (two parallel lines never intercept). You can see this in the graph at the end of the answer.

b) Now we want to find the intercepts algebraically with the equation:

y = (1/3)*x^2

So we write:

(2/3)*x + 5 = y  = (1/3)*x^2

(2/3)*x + 5  = (1/3)*x^2

Now we need to solve this for x.

We can rewrite this as:

-(1/3)*x^2 + (2/3)*x + 5 = 0

For commodity, let's multiply the whole equation by 3, so we get:

3*(-(1/3)*x^2 + (2/3)*x + 5) = 3*0

-x^2 + 2x + 15 = 0

Now we need to find the solutions of that quadratic equation, we can just use the Bhaskara's formula:

$x = \frac{-2 \pm \sqrt{2^2 - 4*(-1)*15} }{2*-1} = \frac{-2 \pm 8 }{-2}$

So we have two solutions for x:

x = (-2 - 8)/-2 = 5

x = (-2 + 8)/-2 = -3

Then we will have two intersections, to get the exact point where the intersection happens we need to evaluate the equation in these two points:

y = (1/3)*5^2 = 25/3

y = (1/3)*(-3)^2 = 3

Then the intersections are at:

(5, 25/3) and (-3, 3)

If you want to learn more, you can read:

https://brainly.com/question/5401922