║×i could really use some the answers are multiple choice as well× ║

the table shows the weight of an animal, in kilograms, over a period of several months, with a question mark for the missing weight.

month: 1, 4, 7, 10, 13.
weight (kg): 40, ? , 62.5, 78, 97.5.

what is most likely the weight of the 4th month
a) 50kg
b) 55.5 kg
c) 65 kg
d) 69.5 kg

I feel like it would be B but I'm not totally sure

See diagram, attached.

Part A: Equation of the directrix is  y= -2

Part B: point of intersection A of directrix & line of symmetry A(6,-2)

Part C: y =  (1/12) (x-6)^2+1

           or on simplifying

            12y = x^2 -12x +48

           (Note: given instructions omitted the +k term.)

part D: The parabola opens upwards becaus p>0.

Part E: The value of p is the difference of y coordinates of F minus the y-coordinate of A, all divided by 2, which is positive.

Part F: p = (4 - (-2))/2 = +3

Part G: y = (1/12)(x-6)^2 + 1

Part H: See attached diagram below

Part I : y = (1/12) (x-6)^2 + 1     where h=6, k=1, or V(6,1).

           It fits with the given formula, provided we include the +k term.

Step-by-step explanation:

The vertex form of the equation of a vertical parabola is given by y= 1/4p(x-h)^2 , where (h, k) is the vertex of the parabola and the absolute value of p is the distance from the vertex to the focus, which is also the distance from the vertex to the directrix. You will use the GeoGebra geometry tool to create a vertical parabola and write the vertex form of its equation. Open GeoGebra, and complete each step below. If you need help, follow these instructions for using GeoGebra.

Part A: Mark the focus of the parabola you are going to create at F(6, 4). Draw a horizontal line that is 6 units below the focus. This line will be the directrix of your parabola. What is the equation of the line? (y=-2)

Part B: Construct the line that is perpendicular to the directrix and passes through the focus. This line will be the axis of symmetry of the parabola. What are the coordinates of the point of intersection, A, of the axis of symmetry and the directrix of the parabola?  A(6,-2)

Part C: Explain how you can locate the vertex, V, of the parabola with the given focus and directrix. Write the coordinates of the vertex.

We know that the equation in vertex form is

y = (1/(4p)(x-h)^2+k   where V(h,k) are the coordinates of the vertex.

From points F and A, we know that the distance from directrix to focus is 6.

Therefore p=3.

The vertex, V(h,k) is located at the mid point between the segment joining the directrix and the focus, namely A & K, which has coordinates V(6,1), thus h=6, k=1.

The equation of the parabola is therefore

y = (1/(4p) * (x-h)^2 +k) = (1/(4*3))(x-6)^2 +1 = (1/12) (x-6)^2+1

(note: given instructions omitted the +k term.)

Simplifying

12y = x^2 -12x +36+12

12y = x^2 -12x +48

Part D: Which way will the parabola open? Explain.

The parabola opens upwards becaus p>0.

Part E: How can you find the value of p? Is the value of p for your parabola positive or negative? Explain.

The value of p is the difference of y coordinates of F minus the y-coordinate of A, all divided by 2, which is positive.

Part F: What is the value of p for your parabola?

p = (4 - (-2))/2 = +3

Part G: Based on your responses to parts C and E above, write the equation of the parabola in vertex form. Show your work.

y = (1/12)(x-6)^2 + 1

Part H: Construct the parabola using the parabola tool in GeoGebra. Take a screenshot of your work, save it, and insert the image below.

See attached diagram below

Part i: Once you have constructed the parabola, use GeoGebra to display its equation. In the space below, rearrange the equation of the parabola shown in GeoGebra, and check whether it matches the equation in the vertex form that you wrote in part G. Show your work.

1.33...x^2 - 16x -16y = -64

Multiply all by 3/4

x^2-12x-12y=-48

rearrange

12y = x^2-12x+36 +12

Factor

12y = (x-6)^2 + 12

Divide by 12 all through

y = (1/12) (x-6)^2 + 1     where h=6, k=1, or V(6,1).

It fits with the given formula, provided we include the +k term.