2.2 KiB
Quadratic Functions
Question 1 a)
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.