# Quadratic Functions In Real-World Context

• conditions for the Quadratic curve to lie completely above or below the x−axis.
• modelling real−world situations using quadratic functions.

## Completing the Square

Question 1:

Express the equation $2x^2 -8x +10$ in the form $a(x-h)^2+k$.

Solution:

We will need to factorise the coefficient of $x^2$ before we can complete the square.

$2x^2-8x+10=2[x^2-4x+5]$

To complete the square, we need to express in the form of $a^2-2ab+b^2$. To do so, we will half the value of the coefficient of $x$ to obtain the value of $b$. To maintain the equality of the expression, we will add another $−2^2$ term to neutralise the term that we have added.

\begin{align} &2[x^2-4x\; \overbrace{+\;2^2-2^2\;+}^0\;5]\\ &\quad^\underbrace{\qquad\qquad\quad}\\ &= 2[(x-2)^2-2^2+5]\\ &= 2[(x-2)^2+1]\\ &=2(x-2)^2+2 \end{align}

## Sketching Graphs of Quadratic Function

Sketching Quadratic Functions in the form $a(x-h)^2+k$.

When we sketch a Quadratic graph for the function $y=a(x-h)^2+k$ with the value of $a<0$  then we have a downward open parabola. You can remember it as the n−shaped graph that has a maximum point. Below is an example of a quadratic graph with $a<0$ with $h=0,k=0$.

However, when we sketch the same graph for a positive value of $a$, where $a=3$, in that case, the graph obtained is an upward open parabola. You can remember it as a u−shaped graph that has a minimum point. Below is an example of a quadratic graph with $a>0$ with $h=0,k=0$.

Let’s consider the u−shaped graph given above. When we change the value of $h$, we see there is a change in the position of the graph along the x−axis. When the value of $h$ is positive the graph shifts towards the positive x−axis or towards the right of the graph.

When the value of $h$ is negative, it shifts towards the negative x−axis which is towards the left of the graph.

However, when the value of $k$ changes it makes the graph shift along the y−axis.

A positive value of $k$ makes the graph shift along the positive y−axis or upwards and a negative value of $k$ shifts the graph towards the negative y−axis or in a downwards direction.

You can summarise the entire point using this image:

Question 2:

Sketch the graph of $y=2(x-1)^2-3$, indicating clearly the coordinates of the turning point and the values where the graph crosses the x & y−axes.

Solution:

Let us look at the information that has been provided to us.

$y=2(x-1)^2-3$

Upon looking at the equation, we get the value of $a=2, h=1$ & $k=-3$.

Looking at the value of $a$, we understand immediately that our graph will be an upward open parabola graph.

The values of $h$ and $k$ show that the graph will shift by 1 point towards the right along the positive $x$−axis and downwards by 3 points along negative $y$ direction. Hence, the minimum point of the graph will naturally be at (1,−3).

To find the $y$−intercept for this we can simply take the equation and substitute $x=0$.

\begin{align}​​ &y =2(x-1)^2-3\\ &y =2(0-1)^2-3\\ &y =2-3\\ &y =-1 \end{align}

Therefore, the y−intercept is  $−1$.

Similarly for x−intercept substituting $y=0$ results in

\begin{align}​​ &y =2(x-1)^2-3\\ &0 =2(x-1)^2-3\\ &3 =2(x-1)^2\\ &1.5 =(x-1)^2 \\ &(x-1)^2 =1.5\\ &(x-1) =\sqrt{1.5}\\ &x = 1 \pm 1.22\\ &x = 2.22 \qquad\text{or}\qquad -0.22 \end{align}

Example 1:

Without sketching the graph, explain whether $y = -2x^2+4x -5$ lies completely above or below the x−axis.

Solution:

First, we need to convert the equation into the square form by completing the square.

\begin{align}​​ &y = -2x^2+4x -5\\ &y = -2[x^2-2x +2.5]\\ &y = -2[x^2-2x+1^2-1^2 +2.5]\\ &y = -2[(x-1)^2-1^2 +2.5] & \text{​(upon dividing coefficient of x by 2)​}\\ &y = -2[(x-1)^2+1.5]\\ &y = -2(x-1)^2-3\\ \end{align}

• Now that we have obtained the square form, we see the shape of the graph as n−shaped since the value of $a=-2$, which is negative.
• Then we find out the maximum point coordinates by looking at the values of $h$ and $k$ obtained when we compare the obtained equation with the standard equation.
• The value of $h=1$ and hence the graph shifts right along $x$−axis by 1 point.
• The value of $k=-3$ and hence the graph shifts down by 3 points.

Since the graph of $y = -2x^2+4x -5$ has maximum point at (1, 3), hence the curve lies completely below the $x$−axis.

Question 3:

Without sketching the graph, explain whether $y=3x^2 +12x +17$ lies completely above or below the x−axis.

Express $y=3x^2 +12x +17$ in the form  $y=a(x+h)^2+k$.

Solution:

\begin{align}​​ &y =3x^2 +12x +17\\ &y =3[x^2 +4x +\frac{17}{3}]\\ &y =3[(x+2)^2-2^2+\frac{17}{3}]\\ &y =3[(x+2)^2+ 1\frac{2}{3}]\\ &y =3(x+2)^2+5\\ \end{align}

Question 4:

Without sketching the graph, explain whether $y=3x^2 +12x +17$ lies completely above or below the $x$−axis. Which of the following statements is true?

1. Since, the graph of \begin{align} y = 3x^2+12x+17 \end{align} has a maximum point of $(-2,5)$, the curve cuts the x-axis at two distinct points.
2. Since, the graph of \begin{align} y = 3x^2+12x+17 \end{align} has a minimum point of $(-2,5)$, the curve lies completely above the x-axis.
3. Since, the graph of \begin{align} y = 3x^2+12x+17 \end{align} has a minimum point of $(-2,5)$, the curve lies completely below the x-axis.
4. Since, the graph of \begin{align} y = 3x^2+12x+17 \end{align} has a maximum point of $(-2,5)$, the curve lies completely below the x-axis.

Solution:

• Since we already have the square form of the equation, $y=3(x+2)^2+5$ from the previous question.
• From the equation, $a=3$ means that the graph is an upward open parabola or u−shaped graph and has a minimum point.
• The $h=-2$, shows the graph is moving 2 units towards the left from origin and the $k=5$ shows the graph is completely above the x−axis and the minimum point of the graph is $(-2,5)$.
• Hence, the graph Option B is correct.

## Modelling real−world situations with Quadratic functions

Quadratic functions are useful for modelling some real−world situations such as projectile motion, shape of objects, & profit functions in economics.

Example 2:

The height from the ground, $h$ metres, of a shuttlecock hit by a boy using a badminton racket can be modelled by the equation $\displaystyle{h=-0.08x^2 + 0.96x +2.12}$ where $x$ is the horizontal distance travelled by the shuttlecock in metres.

1. Find the greatest height reached by the shuttlecock and the corresponding horizontal distance travelled.

Solution:

First, we need to convert the equation into square form,

\begin{align}​​ &h=-0.08x^2 + 0.96x +2.12\\ &h =-0.08[x^2 -12x -26.5]\\ &h =-0.08[(x-6)^2-6^2-26.5] & \text{​(upon dividing coefficient of x by 2)​}\\ &h =-0.08[(x-6)^2 -62.5]\\ &h =-0.08(x-6)^2 +5\\ \end{align}

Since $a=-0.08$, therefore the graph is n−shaped and has a maximum point with the coordinates $(6,5)$.

Thus, the maximum height the shuttlecock reaches is $5 \;m$ and the horizontal distance reached is $6 \;m$.

1. Find the height of the shuttlecock when it first made contact with the racket.

Solution:

Ideally, the shuttlecock would be hit at a height which would be the y−intercept of the shuttlecock’s path.

To find the y−intercept of the quadratic graph we simply substitute $x=0$ in the equation,

\begin{align}​​ &h =-0.08x^2 + 0.96x +2.12\\ &h =-0.08(0)^2 + 0.96(0) +2.12\\ \ &h = 2.12\;m\\ \end{align}

1. The entire length of the badminton court is $13.4 \;m$. Given that the boy stands on one end of the badminton court, will the shuttlecock travel beyond the badminton court? Justify your answer with calculation.

Solution:

When we look at the graph, we understand that the point where the shuttlecock drops is the x−intercept of the graph which can be calculated by substituting $h=0$.

\begin{align*} &h=-0.08x^2 + 0.96x +2.12 \\ &0 =-0.08(x-6)^2 +5 \\ &0.08(x-6)^2 =5 \\ &(x-6)^2 =\frac{5}{0.08} \\ &(x-6) = \pm\sqrt{\frac{5}{0.08}} \\ &(x-6) = \pm\sqrt{62.5} \\ &x = 6\pm {\sqrt{62.5}} \\ &x = 13.9 \qquad \text{or} \qquad -1.91 \end{align*}

Since distance travelled cannot be negative, hence the positive value is taken in consideration.

The length of the court is $13.4 \;m$ and the shuttlecock travels $13.9 \;m$ which is beyond the court.

## Conclusion

In this article, we have learned about the conditions for quadratic curves to lie completely above or below the x−axis. We have also learnt how to complete the square of quadratic function and obtain their Maximum or Minimum point respectively.

We also looked at how to model real−world situations using quadratic functions such as projectile motion. Practising questions related to this concept will help you in developing a good understanding of the topic.

Keep learning! Keep improving!

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