From 10574d8ee4172bdcafbc09f7bc1b59a6d49b73da Mon Sep 17 00:00:00 2001
From: James Su <amagicalsoup@gmail.com>
Date: Mon, 30 Dec 2019 03:00:19 +0000
Subject: [PATCH] Update Quadratic Functions.md

---
 .../Quadratic Functions.md                    | 65 ++++++++++++++++++-
 1 file changed, 64 insertions(+), 1 deletion(-)

diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Functions.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Functions.md
index 02b9f78..b702255 100644
--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Functions.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Functions.md	
@@ -2,4 +2,67 @@
 
 ### Question 1 a)
 
-As $`a`$ varies, the graph stretches when $`a \gt 1 `$ and compresses when $`0 \lt a \lt 1`$
\ No newline at end of file
+As $`a`$ varies, the graph stretches when $`a \gt 1 `$ and compresses when $`0 \lt a \lt 1`$
+
+As $`p`$ varies, the graph moves to either the right (when $`p`$ is negative) or left (when $`p`$ is positive).
+
+As $`q`$ varies, the graph moves either up (when $`q`$ is positive) or down (when $`q`$ is negative).
+
+### Question 1 b)
+
+I would first find the vertex which is equal to is at (AOS, optimal value), or $`(\dfrac{-b}{2a}, \dfrac{-b^2}{4a} + c)`$.
+
+In this case it would be at $`(\dfrac{-13}{6}, \dfrac{-169}{12} + 4)`$
+
+Then by using the step property, which is $`1a, 3a, 5a \cdots \implies 3, 9, 15 \cdots`$, I can plot the points on the graph. In addition, since $`a`$ is positive, the graph
+will be opening upward.
+
+### Question 2 a)
+
+By plugging $`3`$ as the time into the relation $`h = -5t^2 + 100t`$, we get:
+
+$`h = -5(3)^2 + 100(3) \implies h = -5(9) + 300 \implies h = 255`$
+
+The flare will be $`255m`$ tall.
+
+### Question 2 b)
+
+The maximum height reached by the flare is when $`t = \dfrac{-b}{2a}`$ (optimal value).
+
+So, $`\dfrac{-b}{2a} = \dfrac{-100}{-10} = 10`$
+
+$`\therefore h = -5(10)^2 + 100(10) \implies h = 500`$
+
+The maximum height reached by the flare is $`500m`$.
+
+### Question 2 c)
+
+By setting $`h=80`$, we can get the 2 times where the flare reaches $`80m`$, and by taking the difference in $`x`$ values, we get the time the flare stayed above $`80m`$.
+
+$`80 = -5t^2 + 100t`$
+
+$`5t^2 - 100t + 80 = 0`$
+
+$`t^2 - 20t + 16 = 0`$
+
+$`t = \dfrac{20 \pm \sqrt{366}}{2}`$
+
+$`\therefore`$ The duration is $`2(\dfrac{\sqrt{336}}{2}) = \sqrt{336}`$
+
+### Question 3 a)
+
+We can represent the area as $`hw`$, where $`h+w = 20`$, so we can model a quadratic equation as such: $`w(20-w)`$. Therefore the AOS is when $`w=10`$
+
+### Question 3 b)
+
+Since the maximum area is when $`w = 10`$, and $`h = 20 - w \implies h = 10`$. So the dimension is a pen $`10m`$ by $`10m`$.
+
+### Question 4
+
+The cross-sectional area can be modeled by the equation $`(50-2x)x`$. 
+
+Therefore the AOS is when $`\dfrac{25}{2}`$ since $`x=0, 25`$ are the solutions to this quadratic equation when it equals $`0`$, and the AOS is the average of them both.
+
+Therefore the value of $`x=12.5cm`$ gives the maximum area for the sectional area.
+
+