Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples on Quadratic Functions

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

Solve all questions.
Show all work.
Pre-requisite: Quadratic Equations

There are two methods that can be used to solve any extrema (maxima or minima) problem.
They are the: Vertex Formula and the Differential Calculus method.
Except for some applicable questions, we shall use the Vertex Formula for the questions.
This is because it is the concept being taught in this section.
If you want to learn the Differential Calculus method in solving extrema problems, please review Differential Calculus

(1.) For the function: $y = -2x^2 + 8x + 11$
(a.) Determine the vertex.
(b.) Write the function in Vertex form.
(c.) Determine the $y-intercept$
(d.) Determine the $x-intercept$


$ y = -2x^2 + 8x + 11 \\[3ex] y = ax^2 + bx + c \\[3ex] a = -2, b = 8, c = 11 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] -\dfrac{b}{2a} = -\dfrac{8}{2(-2)} \\[5ex] -\dfrac{b}{2a} = \dfrac{-8}{-4} = 2 \\[5ex] f\left(-\dfrac{b}{2a}\right) = f(2) \\[5ex] But\:\: f(x) = -2x^2 + 8x + 11 \\[3ex] f(2) = -2(2)^2 + 8(2) + 11 \\[3ex] f(2) = -2(4) + 16 + 11 \\[3ex] f(2) = -8 + 16 + 11 \\[3ex] f(2) = 19 \\[3ex] (a.)\:\: Vertex = (2, 19) \\[3ex] Vertex\:\: Form:\:\: y = a(x - h)^2 + k \\[3ex] Vertex = (2, 19) \\[3ex] Vertex = (h, k) \\[3ex] h = 2, k = 19 \\[3ex] (b.)\:\: Vertex\:\: Form:\:\: y = -2(x - 2)^2 + 19 \\[3ex] $ To find the $y-intercept$, set $x = 0$ and solve for $y$

$ y = -2x^2 + 8x + 11 \\[3ex] x = 0 \\[3ex] y = -2(0)^2 + 8(0) + 11 \\[3ex] y = -2(0) + 0 + 11 \\[3ex] y = 0 + 0 + 11 = 11 \\[3ex] (c.)\:\: y-intercept = (0, 11) \\[3ex] $ To find the $x-intercept$, set $y = 0$ and solve for $x$

$ y = -2x^2 + 8x + 11 \\[3ex] y = 0 \\[3ex] 0 = -2x^2 + 8x + 11 \\[3ex] -2x^2 + 8x + 11 = 0 \\[3ex] a = -2, b = 8, c = 11 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-8 \pm \sqrt{(-8)^2 - 4(-2)(11)}}{2(-2)} \\[5ex] x = \dfrac{-8 \pm \sqrt{64 + 88}}{-4} \\[5ex] x = \dfrac{-8 \pm \sqrt{152}}{-4} \\[5ex] x = \dfrac{-8 \pm \sqrt{4 * 38}}{-4} \\[5ex] x = \dfrac{-8 \pm \sqrt{4} * \sqrt{38}}{-4} \\[5ex] x = \dfrac{-8 \pm 2 * \sqrt{38}}{-4} \\[5ex] x = \dfrac{-2\left(4 \mp \sqrt{38}\right)}{-4} \\[5ex] x = \dfrac{4 \mp \sqrt{38}}{2} \\[5ex] x = \dfrac{4 - \sqrt{38}}{2} \:\:\:OR\:\:\: x = \dfrac{4 + \sqrt{38}}{2} \\[5ex] (d.)\:\: x-intercepts = \left(\dfrac{4 - \sqrt{38}}{2}, 0\right) \:\:and\:\: \left(\dfrac{4 + \sqrt{38}}{2}, 0\right) $
(2.) ACT In the standard $(x, y)$ coordinate plane, the graph of $y = 30(x + 17)^2 - 42$ is a parabola.
What are the coordinates of the vertex of the parabola?

$ A.\:\: (-30, -42) \\[3ex] B.\:\: (-17, -42) \\[3ex] C.\:\: (17, -42) \\[3ex] D.\:\: (17, 42) \\[3ex] E.\:\: (30, 42) \\[3ex] $

$ y = 30(x + 17)^2 - 42 \\[3ex] Compare to \\[3ex] Vertex\:\:Form\:\:of\:\:Quadratic\:\:Function \\[3ex] y = a(x - h)^2 + k \\[3ex] -h = 17 \\[3ex] \rightarrow h = -17 \\[3ex] k = -42 \\[3ex] Vertex = (h, k) \\[3ex] \therefore Vertex = (-17, -42) $
(3.)


We can do this question in two ways.
You can use any of the methods to solve it.

First Method: Vertex Formula - because the profit is a quadratic function
For this question:
the number of bags that will give the maximum profit = $x$ coordinate of the vertex
the maximum profit = $y$ coordinate of the vertex

$ y = 10x - x^2 \\[3ex] y = -x^2 + 10x \\[3ex] y = ax^2 + bx + c \\[3ex] a = -1, b = 10 \\[3ex] x-coordinate\:\: of\:\: Vertex = -\dfrac{b}{2a} \\[5ex] = \dfrac{-10}{2(-1)} \\[5ex] = \dfrac{-10}{-2} \\[5ex] = 5 \\[3ex] $ The sale of $5$ bags will give the maximum profit

Second Method: Differential Calculus
One of the application of derivatives is in calculating maxima and minima of functions
You need to review Differential Calculus before using this method.

$ y = 10x - x^2 \\[3ex] \dfrac{dy}{dx} = 10 - 2x \\[3ex] 10 - 2x = 0 \\[3ex] 10 = 2x \\[3ex] 2x = 10 \\[3ex] x = \dfrac{10}{2} \\[3ex] x = 5 \\[3ex] $ The sale of $5$ bags will give the maximum profit
(4.)


Normal revenue for $1$ week = $40 * 22 = \$880$
But, the store wants to make more money.
For every $\$1$ decrease in price, the store will sell $4$ more caps per week.
For $22 - 1 = 21$, the store will sell $40 + 4 = 44$ caps
The company will sell $4$ more caps for a $\$1$ decrease in price
Revenue would be $21 * 44 = \$924$
For $22 - 1(2) = 20$, the store will sell $40 + 4(2) = 40 + 8 = 48$ caps
The company will sell $8$ more caps for a $\$2$ decrease in price
Revenue would be $20 * 48 = \$960$
For $22 - 1(3) = 19$, the store will sell $40 + 4(3) = 40 + 12 = 52$ caps
The company will sell $12$ more caps for a $\$3$ decrease in price
Revenue would be $19 * 52 = \$988$

Student: That would exactly take more than a minute to solve.
ACT questions typically should be solved at at an average of $1$ question per minute.
Teacher: I expected that comment.
I did those steps to explain the question to you.
I was doing it the Arithmetic method.
Let us now do it algebraically.


So;
For each $\$x$ decrease in price, the company will sell $4x$ more
For $22 - 1(x) = 22 - x$, the store will sell $40 + 4(x) = 40 + 4x$ caps
Revenue would be $(22 - x) * (40 + 4x)$
Let Revenue = $R$

$ R = (22 - x)(40 + 4x) \\[3ex] = 880 + 88x - 40x - 4x^2 \\[3ex] = 880 + 48x - 4x^2 \\[3ex] = -4x^2 + 48x + 880 \\[3ex] R = -4x^2 + 48x + 880 \\[3ex] y = ax^2 + bx + c \\[3ex] a = -4, b = 48 \\[3ex] x-coordinate\:\: of\:\: vertex = -\dfrac{b}{2a} \\[5ex] = -\dfrac{48}{2(-4)} = \dfrac{-48}{-8} = 6 \\[5ex] $ The store has to decrease the price by $\$6$
This means that the store has to sell each cap for $22 - 6 = \$16$
By doing so, the store will sell $40 + 4(6) = 40 + 24 = 64$ caps
Revenue = $16 * 64 = \$1024$

Student: So, this is the maximum revenue the store can get?
Teacher: That is correct.
Student: What if the store sells each cap for $\$15$?
Teacher: Go ahead and find the revenue
Student: Okay...it will be
$15 * (64 + 4) = 15 * 68 = \$1020$
Interesting application of Quadratic Functions!