graphing inverse

Question

with the given points (4,2) (2,-1) (-2,-2) find the reflection of the original graph.

Answer 1

Hi Tim, reflected equations and inverse equations are not the same thing. The question asks for the reflection of the graph. That part will be easy once you have the original equation. For the original equation you will need to solve a three equation system of quadratic equations. That part is not so easy, but once you have that all you need to do is change the sign of the x² term. That reflects the entire graph across the x-axis without changing the shape of the graph.

Solving for the original equation is a multistep proces, here goes:

Each of the three coordinates are points that must be on the graph of the equation you are looking for. So you use each coordinate to generate three equivalent quadratic equations, then solve them to find the coefficients.First, use the coordinates you are given to substitute for x and y to get three quadratic equations in the form of y = ax2 + bx + c.

1. coordinates (4,2), x = 4, and y = 2

2y = a(4²) + 4b + c, or 2y = 16a + 4b + c

2, coordinates (2, -1), x = 2, and y = -1

-y = a(2²) + 2b + c, or -y = 4a + 2b + c

3.coordinates ((-2,-2), x = -2, and y = -2)

-2y = a (-2²) -2b + c, or -2y = 4a - 2b + c.

Now that you have three equations, the next step is to subtract equations to eliminate one of the variables, resulting in two equations with only two variables. For this problem, let's subtract equation 2 from equation 1, and subtract equation 3 from equation 2. That will eliminate the coefficient c.

Eq. 1 minus eq. 2:

2 = 16a + 4b + c

-1 = 4a + 2b + c

-________________

3 = 12a + 2b (call this equation 4)

Eq 2. minu2 eq. 3

2 = 16a + 4b + c

-2 = 4a - 2b + c

-________________

4 = 12a + 6b (call this equation 5)

So equations 4 and 5 give us a two variable, two equation system that can also be solved for the variables by subtraction. Let's subtract equation 4 from equation 5 to eliminate the a term, then all that will be left is b and we can solve for that.

4 = 12a +6b

3 = 12a + 2b

______________

1 = 4b

1/4 = b

Now that b is solved, we can plug that back into either equation 4 or 5 and solve for the a term. Let's use equation 5:

3 = 12a + 2(1/4)

3 = 12a + 1/2

2 1/2 = 12 a

5/2 = 12 a

5/24 = a

Now that we know the values of a and b, those can be plugged into any of equations 1, 2 or 3 to solve for c. For equation 1 that would be:

2 = 16a + 4b + c

2 = 16(5/24) + 4(1/4) + c

2 = 80/24 + 1 + c

2 = 104/24 + c

2 = 13/3 + c

6/3 = 13/3 + c

-7/3 = c

If this is correct, then we should get the same value for c using equations 2 or 3. Let's check:

Equation 2:

-1 = 4a +2b + c

-1 = 4(5/24) + 2(1/4) + c

-1 = 20/24 + 1/2 + c

-1 = 20/24 + 12/24 + c

-1 = 32/24 + c

-1 = 1 1/3 + c

-2 2/3 = c

-7/3 = c, check, so far so good, now for equation 3:

-2 = 4a -2b + c

-2 = 4(5/24) -2(1/4) + c

-2 = 20/24) -1/2 + c

-2 = 20/24 -12/24 + c

-2 = 8/24 + c

-2 = 1/3 + c

-2 1/3 = c

-7/3 = c. Perfect! All three equations give the same value for c, so now we know the cofficients for the quadratic equation. Those are:

a = 5/24

b = 1/4

c = - 7/3

The quadratic then is:

y = 5/24x² + 1/4b -7/3, and the reflected equation is:

y = -5/24x² + 1/4b -7/3

To eliminate fractions in the coefficients, multiply both sides of the equations by 24 giving:

24y = 5x² + 6b - 56, and reflected:

24y = -5x² + 6b -56.

It's a long process, but I hope that helps