I meant exactly what I said: solve the equation for y! If I gave you the equation x^2=-28(y-8.5) and asked you to solve for y, what would you do? (Some people call it "making y the subject".)

Hi Dr.Peterson, thanks for getting back to me. I appreciate the time you are taking to help me out.

So far on parabolas I have learnt how to identify them using the key information (vertex, focus, centre, directrix and latus rectum ends) and drawing them onto graphs. I have also learnt about parametric equations, implicit differentiation, gradients, tangents, normals and intersections.

Coming across this exercise in the study has put me to a halt. I am unfamiliar with this type of parabola and problem. The only exercise I have done was this one:

What do you mean when you say 'find the y'? I have not come across an exercise like that in this study before. I have with ellipses and circles but not parabolas. Do you mean I must differentiate the parabola equation?

Why would "a" be half of the base measurement? The example you think you are following is entirely different (it doesn't mention the focus), and it doesn't set a to half the base width, but instead replaces x with that number and solves for a.

I don't understand when you say this example is entirely different. Are you saying I should be using a different technique??

The example is the only method I have learnt in regards to problem solving. I will explain how I got the idea...

EXAMPLE states:

*- HEIGHT of 2.4metres*

- WIDTH of 1.6metres
If you divide the width by half you get 0.8metres

It was also stated in the example that the parabola intersects the x-axis at (8,0) and (-8,0).

0.8metres is exactly what I got by dividing the width.

The form used:

**y= ax^2+b**
for

**'a'** they used

**0.8metres **(width from centre to end of parabola)

for

**'b'** they used

**2.4metres** (height from centre to top of parabola)

If I use the same form but give different figures using the design I am working on:

- HEIGHT of 8.5metres

- WIDTH of 36metres

Divide the width by half you get 18metres

It was also stated that whether or not the width of the pool roof is less the 36 metres - I wouldn't have a clue what the width is, since 36metres was mentioned in the first place, I assume that number is what the width is.

a= 18metres

b= 8.5metres

But you didn't even do what you say you're doing; you replaced "ax^2" with "18^2", not with 18x^2.

Can you explain please how it makes any difference adding an 'x' in there?

I shall correct it then...

**y= 18x^2 + 8.5**
sorry if this has confused you, but I really need an explanation for everything.

The height above the pool at any point means the value of y for any x within the pool (that is, between -9 and 9). As you can see from the graph, or just know about quadratic equations with negative leading coefficient, y decreases as you move away from the axis, so you only need to check the value of y when x = 9 (or -9), which are the locations of the ends of the pool.

THANK YOU!! That makes total sense! I wondered maybe those particular points could be used since it was the only information given other than the focus and vertex. I just don't know how to solve this now.

I don't know what I am doing so I went online to use the symbolab calculator which hopefully has given me a correct answer:

Y^2=-28(y-8.5)

EXPAND

(x^2)(-28y)((-28x-8.5=238)

x^2=-28y+238

SIMPLIFY

x^2= 28-238

x^2= 210

x^2= √210 = 14.491

x= 14.5

I am hoping this is the answer to the distance from the centre to the x-intercept either side?

So if I multiply it by 2, the total distance is 29metres?

I am still not sure where to go from there. I feel that somehow I need to substitute Point (9,0) into the equation but I don't know how to do it.

Can someone just please make it clear to me because I really don't know what to do. I'm just guessing everything now and I don't know where I'm going and I would like some guidance for the answer so in the future I can remember and use it again. Please, thank you