cadaVR anatomy by Mozaik Education

The quadratic function

Henceforth we deal with functions where the largest appearing index of the variable \latex{ x } is the second index. These functions are called quadratic functions.

Example 1

The area of a square with side \latex{ a } \latex{(\text{e.g.}\,cm)} is \latex{ a^{2} } \latex{ (cm^{2}) }. In connection with this we can interpret function \latex{ g(a) = a^{2} } on the positive numbers. Let us plot the graph of function \latex{ f } which is an extension of this function to all real numbers, i.e. the following function:

\latex{f:\R\rightarrow \R}, \latex{f(x)=x^{2}}.

Solution

The image of function \latex{ f } is a nice curve which is called parabola (Figure 33). In geometry later on we are going to learn that a parabola is the set of points in the plane which are equidistant from a given straight line \latex{e} and from a given point \latex{ F } which does not lie on straight line \latex{ e } (Figure 34). Point \latex{ F } is the focus of the parabola; straight line \latex{ e } is the directrix of the parabola. The axis of the parabola received as the graph of function \latex{ f } is the \latex{ y }-axis, its vertex is the origin.

Let us characterise the function f : \latex{\R\rightarrow\R,} \latex{f (x) = x^{2}}.

It is decreasing on \latex{]–\infty; 0]}, it is increasing on \latex{[0; +\infty[}, it has a minimum at \latex{ x=0 }, its value is \latex{f(0)=0}. It can be proved that the range of the function is the interval \latex{[0; +\infty[ }, i.e. the set of non-negative real numbers.

The graph of the function, the parabola is symmetric about the \latex{ y }-axis. The reason is simple, since \latex{(-x)^{2}=x^{2}}, so if a point \latex{P(x; y)} lies on the parabola, then the image \latex{P’(–x; y)} of point \latex{ P } when reflected about the \latex{ y }-axis also lies on the curve. This property of function \latex{ f } can also be formulated as follows: \latex{f(–x) = f (x)}.

DEFINITION: The functions, where \latex{f(–x)=f(x)} is fulfilled for every x of the domain, are called even functions.

For example \latex{g:\R\rightarrow\R}, \latex{g(x)=\mid x\mid} is an even function, since it is true that \latex{g(–x)=g(x)} is fulfilled for every element of the domain. The graph of function \latex{ g } is also symmetric about the \latex{ y }-axis.

Example 2

Let us plot the graph of the following \latex{\R\rightarrow\R} type functions:

\latex{g(x) = (x - 2)^{2}}; \latex{h(x) = x^{2} - 4}.

Solution

If we create the table of values to plot the graph of function \latex{ g }, we shortly realise that \latex{ (x-2)^{2} } takes the same values at a place \latex{ 2 } greater than \latex{ x^{2} } does.

For example it takes \latex{ 0 } at \latex{ 2;\, 1 } at \latex{ 3 }; and \latex{ 4 } at \latex{ 2 }. When plotting the graph it means that by translating the image of \latex{ x^{2} } by \latex{ 2 } into the positive direction along the \latex{x}-axis we get the image of \latex{ (x-2)^{2} }.
It is even more obvious from the definition of function \latex{ h } that we get the graph of \latex{x^{2} - 4} from the graph of \latex{ x^{2} } by translating it by \latex{ 4 } units into the negative direction along the \latex{ y }-axis. (Figure 35)

Example 3

Let us plot the graphs of and characterise the following \latex{\R\rightarrow\R} type functions:

\latex{f(x) = x^{2} - 2x + 3};

\latex{g(x)=1-x^{2}};

\latex{h(x)=-x^{2}+4x-3}.

Solution

Function \latex{ f } can also be formulated as follows, i.e. by completing the square:

\latex{f (x) = (x^{2} - 2x + 1) + 2 = (x - 1)^{2} + 2}.

So the parabola, which is the image of function \latex{ x^{2} }, should be translated by \latex{ 1 } to the right and by \latex{ 2 } upwards (into the positive direction). (Figure 36)

So function \latex{ f } is decreasing on the interval \latex{]–\infty; 1]}, it is increasing on \latex{[1; +\infty[;} it has a minimum at \latex{x = 1}. Its minimum value is \latex{f (1) = 2}. The range of \latex{ f } is interval \latex{[2; +\infty[}.

Let us rewrite function \latex{g} as follows: \latex{g(x) = -x^{2} + 1}, and let us first plot the graph of function \latex{x\mapsto-x^{2}}. It differs from function \latex{ x^{2} } in a way that its value is \latex{ –1 } times the value of the previous at all places, which means that the image of \latex{ x^{2} } should be reflected about the \latex{ x }-axis. The image of \latex{ g } is derived from this by translating it by \latex{ 1 } unit upwards. (Figure 37)

It is practical to rewrite the function \latex{ h } as follows:

\latex{h(x) = -(x^{2} - 4x + 3) = -(x - 2)^{2} + 1}.

When plotting the image of \latex{ h } using this form we get the graph shown in Figure 38. Function \latex{ h } is increasing on the interval \latex{]–\infty ; 2]}, it is decreasing on the interval \latex{[2; +\infty [}; it has a maximum at \latex{ 2 }. Its maximum value is \latex{h(2) = 1}. The range of \latex{ h } is interval \latex{]–\infty ; 1]}.

Figure 36

\latex{y = x^2 - 2x + 3}

\latex{y=x^{2}}

\latex{y=(x-1)^{2}}

\latex{ 2 }

\latex{ 1 }

\latex{ y }

\latex{ x }

\latex{ 1 }

Figure 37

\latex{y=-x^{2}}

\latex{y=1-x^{2}}

\latex{y=x^{2}}

\latex{ -1 }

\latex{ 1 }

\latex{ y}

\latex{ x}

Figure 38

\latex{y=-x^{2}+4x-3}

\latex{y=x^{2}}

\latex{y=(x-2)^{2}}

\latex{y=-(x-2)^{2}}

\latex{ 1 }

\latex{ 2 }

\latex{ 3 }

\latex{- 3 }

\latex{y }

\latex{x }

\latex{ 1 }

Example 4

Using the knowledge obtained about the functions let us solve the following exercise. Two ships are heading along two perpendicular straight lines towards the intersection point of the straight lines in the sea. At a given time point the first ship, which is cruising at a speed of \latex{ 30 } \latex{ \frac{km}{h} } , is \latex{ 100 } \latex{ km } away from intersection point \latex{ O } of the two straight lines. At the same time the second ship, which is cruising at a speed of \latex{ 40 } \latex{ \frac{km}{h} } , is \latex{ 300 } \latex{ km } away from point \latex{ O }. At what time will the two ships be the closest to each other and what is this shortest distance?

Solution

Let us denote the time measured in \latex{ hours } and elapsed since the given time point by \latex{ t }, and let us calculate the distance of the two ships at time point \latex{ t }.

The first ship, the place of which at time point \latex{ t } is denoted by \latex{ P }, has covered \latex{ 30t } \latex{ kilometres }, thus its distance measured from \latex{ O } is: \latex{PO = 100 - 30t}. If point \latex{ P } was to the right of point \latex{ O }, then this distance would be \latex{ 30t - 100 }, thus the formula for the distance which is correct in both cases: \latex{PO =\mid100 - 30t\mid}. (Figure 39)

The distance of the second ship denoted by \latex{ Q } from point \latex{ O } is similarly \latex{QO =\mid300 - 40t\mid}. From the right-angled triangle \latex{ POQ } using the Pythagorean theorem the square of distance \latex{ PQ } is:

\latex{PQ^2 = (100 - 30t)^{2} + (300 - 40t)^{2}}.

Obviously the square of the distance is the shortest exactly when the distance is the shortest. So the exercise can be formulated as follows: the function

\latex{f (t) = (100 - 30t)^2 + (300 - 40t)^2,} \latex{t\geq0}

is given, let us find its minimum place and its minimum value.

Let us first transform the expression defining the function:

\latex{(100-30t)^{2}+(300-40t)^{2}=}
\latex{=10,000-6,000t+900t^{2}+90,000-24,000t+1,600t^{2}=2,500(t^{2}-12t+40)}.

So it is enough to find the minimum place of function

\latex{g(t) = t^{2} - 12t + 40 = (t - 6)^{2} + 4,} \latex{t\geq0},

the value of f will be the smallest at the same place.

Let us plot the graph of function \latex{ g } (Figure 40). The image of function \latex{ g } is a parabola, which we get from the so called normal parabola \latex{(}from the image of function \latex{x^{2})} by translating it by \latex{ 6 } to the right and by \latex{ 4 } upwards. The minimum place of \latex{ g } is at \latex{ t = 6 }, its minimum value is \latex{ g(6) = 4 }. According to this the minimum place of\latex{ f } is also at \latex{ t = 6 }, its minimum value is \latex{f (6) = 2,500 × 4 = 100^2}.

So the answer to the question of the exercise is as follows: the two ships will be closest to each other in \latex{ 6 } \latex{ hours } measured from the given time point, their distance will be \latex{ 100 } \latex{ km }.

We have already drawn the graph of function \latex{ x^{2} } , the (normal) parabola several times. We always draw it in a way that if we take the chord connecting two points of the parabola, then the curve between the two points will be below the chord (Figure 41). Is it really so?

We are going to show now that if \latex{ a\lt b }, and if we draw the chord connecting points \latex{(a; a^2)}, \latex{(b; b^2)}, then the point of the parabola at the mind-point of interval \latex{\left[a;b\right] } will indeed be below the chord. This property of the curve of function \latex{ x^{2} } is expressed as the curve (from below) is convex. (Figure 42)

Using the notations of the figure it is enough to show that \latex{ FR\lt FS }.
The length of line segment \latex{ FR } is: \latex{\Biggl(\frac{a+b}{2}\Biggr)^{2}}.
Line segment \latex{ FS } is the mid-line of trapezium \latex{ ABQP }, thus \latex{F=\frac{PA+QB}{2}}.

Since \latex{PA = a^{2}}, \latex{QB = b^{2}}, it is enough to show that \latex{\Biggl(\frac{a+b}{2}\Biggr)^{2}\lt \frac{a^{2}+b^{2}}{2}}.

Let us do the squaring on the left side, and let us multiply both sides of the inequality
by \latex{ 4 }:

\latex{a^{2} + 2ab + b^{2}\lt2a^{2} + 2b^{2}.}

The right side is greater than the left side if and only if the difference of the right side and the left side is positive. And it is true, because the difference is:

\latex{a^{2}-2ab+b^{2}=(b-a)^{2}\gt0},

since \latex{ b\gt a }. Since the steps can be reversed, the original statement is also true.

Exercises

{{exercise_number}}. Let us plot the graphs of and characterise the following \latex{\R\rightarrow \R} type functions.

\latex{f (x) = x^{2} + 1}

\latex{g(x) = -x^{2}}

\latex{h(x) = (x-1)^{2}}

\latex{k(x) = -(x + 1)^{2}}

\latex{l(x) = -x^{2} + 4}

{{exercise_number}}. Let us plot the graphs of and characterise the following \latex{\R\rightarrow \R} type functions.

\latex{f (x) = 2x^{2}}

\latex{g(x)=\frac{1}{2} x^{2}}

\latex{h(x) = x^{2}-6x + 5}

\latex{k(x) = -x^{2}-4x + 2}

{{exercise_number}}. We throw a stone vertically upward with a velocity of \latex{ 20 } \latex{ \frac{m}{s} } . What distance does it go upward and after what time does it fall back to the ground? (Let us choose \latex{ 10 } \latex{ \frac{m}{s^{2} } } rounded as the acceleration due to gravity, and let us neglect air friction. The stone has two different displacements: it can be calculated as the resultant of a steady motion upward and a steady motion downward with a constant acceleration. The function of displacement of this latter one is \latex{s(t)=-\frac{1}{2} gt^{2}} . )

Mathematics 9.

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