1 (a) A function passes through the points (0, -5), (1, 0), (2, 7). Use finite differences to

determine the equation of the function.
(b) Draw the graph of the function.
(c) Draw the inverse on the graph.
(d) Show at least two different restrictions to the domain that could be made so that the
inverse is a function.
(e) Find the equation of the inverse.
(f) Show using the domains found in part (d) that the domain of the function is the
range of the inverse and that the range of the function is the domain of the inverse.

What is your question about this assignment?

the whole thing

if u know da workin out

With only those three points, the only definite thing we can say,

it is not a linear function.

With three points, I could find a quadratic function, or a cubic function.
I could even find the equation of a circle passing through those points.

To have a quadratic function, the second differences have to be all the same constant, but we only get a single value of 2, so it is not enough information.

(a) To determine the equation of the function using finite differences, we need to find the differences between consecutive y-values for each set of x-values. Let's start by calculating the differences for the given points:

(0, -5), (1, 0) --> Δy = 0 - (-5) = 5
(1, 0), (2, 7) --> Δy = 7 - 0 = 7

Now, we calculate the second differences:

Δ²y = 7 - 5 = 2

Since the second differences are constant, we can conclude that the function is a quadratic equation in the form of y = ax² + bx + c. We can substitute one of the points into this equation to solve for the coefficients a, b, and c.

Using the point (0, -5):
-5 = a(0)² + b(0) + c
-5 = c

So, we have c = -5. Now, substituting another point, say (1, 0), we get:

0 = a(1)² + b(1) - 5
0 = a + b - 5

We now have two equations:

Equation 1: a + b = 5
Equation 2: c = -5

The system of equations can be solved to find the values of a and b. Adding equation 1 and equation 2, we get:

a + b = 5
c = -5
---------
a + b + c = 0

Substituting c = -5 into the equation, we have:

a + b - 5 = 0
a + b = 5

Since we have the same equation in both equation 1 and the modified equation, we can conclude that a = 0. Substituting a = 0 into equation 1, we find:

0 + b = 5
b = 5

Therefore, the equation of the function is y = 0x² + 5x - 5, which simplifies to y = 5x - 5.

(b) To draw the graph of the function, plot the given points (0, -5), (1, 0), and (2, 7) on a coordinate plane. Then, connect the points with a smooth curve since the function is quadratic. The graph should resemble a U-shaped curve with the vertex at the turning point.

(c) To draw the inverse on the graph, we need to reflect the points over the line y = x. Switch the x and y coordinates of the original points and plot them accordingly. Connect these points with a curve. The graph of the inverse should be the reflection of the original function's graph across the line y = x.

(d) Two different restrictions we could make to ensure that the inverse is a function are:
1. Restrict the domain to only include the portion of the graph that passes the vertical line test. This means excluding any vertical stretches or compressions of the graph that could potentially create more than one output for a given input.
2. Restrict the domain to a specific range of x-values where the function is one-to-one, meaning that each x-value corresponds to exactly one y-value and vice versa.

(e) To find the equation of the inverse, we interchange the x and y variables in the original function equation and solve for y:

x = 5y - 5
x + 5 = 5y
y = (x + 5) / 5

So, the equation of the inverse function is y = (x + 5) / 5.

(f) Let's examine the domains and ranges found in part (d) to show the relationship between the function and its inverse:

If we restrict the domain of the function to be the range where it passes the vertical line test, this would become the range of the inverse. Similarly, if we restrict the domain of the inverse to be a specific range of x-values where it is one-to-one, then this would become the domain of the original function.

By restricting the domains as described above, we ensure that each function is defined within its range and inverse domain, respectively. The range of the original function is then equal to the domain of its inverse, while the domain of the original function is equal to the range of its inverse.