Week -3 | System of Equations - 2

1.1 Graph of a Quadratic equation

The graph represened by $y=ax^2+bx+c$ is a parabola (You can read about it more by going to the link) for all real triplets $(a,b,c)$. While the shape of the graph remains the same, the nature depends on the value of this triplet. Consider the following graph:

Fig 1.1.1

This graph intersects the X-Axis at points $(-2,0)$ & $(1,0)$. Since we know that $y=0$ for all points on the X-Axis, we can say that $1$ & $-2$ are roots of this graph. The graph defined above is a parabola. Any parabolic curve has the form of a quadratic equation, thus we know the equation of this curve is $y=a(x+2)(x-1)$.

Consider this graph shown below:

Fig 1.1.2

Fig 1.1.2 also represents a curve of the form $y=a_0(x+2)(x-1)$. But do you observe the difference between these 2 curves? The second curve is much steeper than the first curve. In fact, the first figure is for $y=(x+2)(x-1)$ while the second curve is for the equation $y=5(x+2)(x-1)$.

Now consider the following graph:

Fig 1.1.3

This is the graph of the equation $y=-(x+2)(x-1)$. If we compare Fig 1.1.3 & Fig 1.1.1 we must note that the former is a reflection of the latter about the X-axis. We must note that thus, the graph of $g(x)=-f(x)$ is the reflection of the curve of $f(x)$ about the X-Axis. Thus for any x_0, $(x_0, f(x_0))$ must lie on the graph of $f(x)$. In the case of quadratic expressions, $f(x)=ax^2+bx+c$ for some real triplet $(a.b.c)$.

Recall from our earlier chapters the function $f(x)=|x|$, which, for any real $x$ denotes the magnitude of $x$. What about the graph of $y=|(x+2)(x-1)|$?

To answer this, let us consider the basic function $f(x)=|x|$.

$f(x) =x$, $x \ge 0$ and

$f(x)=-x$, $x<0$

Fig 1.1.4

This is the graph of $y=|x|$. Note that for $x>0$, we have the graph of $y=|x|$ resemble the graph of $y=x$. For $x<0$, $|x|=-x$, thus the graph is the reflection of $y=x$ about the X-Axis.

Let’s consider Fig 1.1.1 which represents the graph of $y=f(x)=(x+2)(x-1)$. We have $f(-2)=f(1)=0$. What happens when $-2<x<1$? Without any calculations, let’s note from the graph that $f(x)<0$ when $-2<x<1$. Thus, to find the graph of $|f(x)|$, we need to flip, or reflect $f(x)$ for all $x \in (-2,1)$ about the X-Axis. This gives us the following graph for $y=|(x+2)(x-1)|$:

Fig 1.1.5

Test Your Concepts:

Draw the graph of $f(x)=|x|^2-10|x|+20$. How many real values of $x$ satisfy $f(x)=0$?

1.2 Range of a Quadratic Equation

We’ll consider parabolas more elaborately in our Coordinate Geometry module. For now, we must note that a parabola is symmetric about its axis. An axis passes through the parabola’s vertex dividing the curve into 2 symmetric halves. Vertex of a parabola is defined to be the point where it changes direction. Thus, to find the axis of a parabola, we need to draw a line through the vertex that divides the parabola into identical halves. For now, let’s assume that a parabola of the form $ax^2+bx+c$ has its axis parallel to the Y-Axis always.

Fig 1.2.1

For the function $f(x)=(x+2)(x-1)$, the point $B$ marks the vertex of the parabola. $B$, in this case marks the minimum value of the quadratic expression $(x+2)(x-1)$. Note that a line parallel to the Y-Axis and passing through $B$ should thus divide the parabola into identical halves. The Line $x=-\frac{1}{2}$ thus divides the parabola symmetrically.

Let’s try to find the value $x_0$ which wil be the axis for a quadratic $ax^2+bx+c$. Consider Fig 1.2.1, in this case we must observe that the roots of the equation should be equidistant from the intersection of the X-Axis and the parabola axis line (since the parabola is symmetric about its axis). We know the roots of this equation are $x=-2$ & $x=1$. Thus the intersection point of the parabola’s axis and the X-Axis should be equidistant from these 2 points and hence, we conclude that the line should be $x=\frac{-2+1}{2}=-\frac{1}{2}$. What about any general parabola with roots $\alpha$ & $\beta$? The axis line thus should be $x=\frac{\alpha + \beta}{2}$.

We also know that $\alpha + \beta = -\frac{b}{a}$, from which we conclude that for any general quadratic of the form $ax^2+bx+c$, the axis is the line $x=-\frac{b}{2a}$. Suppose that $a>0$, when $x \rightarrow \infty$, we can have $ax^2>bx$, since $x^2$ shall have a far greater value than $x$. In fact, consider this statement : For any positive $a$ and any $b$, there exists a positive real $M$ such that for all $x>M$, $ax^2>bx$. Thus we note that for $a>0$, $ax^2+bx+c \rightarrow \infty$ as $x \rightarrow \infty$. We thus conclude that there is no upper bound on $f(x)=ax^2+bx+c$ for $a>0$.

$ax^2+bx+c = a\left[x^2+\frac{b}{a}x +\frac{c}{a} \right ]$ $=a\left[x^2+\frac{b}{a}x +\frac{c}{a}+\frac{b^2}{4a^2} - \frac{b^2}{4a^2} \right ] = a\left[ \left(x+\frac{b}{2a} \right ) ^2 +\frac{c}{a}-\frac{b^2}{4a^2}\right]$

We can thus conclude that for $a>0$, the minimum value of the quadratic expression appears at $x=-\frac{b}{2a}$. Similarly for $a<0$, we have $ax^2+bx+c \rightarrow – \infty$ as $x \rightarrow \infty$ and hence it does not have a lower bound. Thus $a$, which is called the leading coefficient of thequadratic expression, determines the range of values that the expression can take. Also noteworthy is the fact that for negative leading coefficient, the maximum value occurs at $x=-\frac{b}{2a}$. We can thus say that the extremum of the quadratic expression $y=ax^2+bx+c$ occurs at $x=-\frac{b}{2a}$.

2.1 Inequations

Some notations :

$a\le x <b \iff x \in [a,b)$

$a< x <b \iff x \in (a,b)$

$a\le x \le b \iff x \in [a,b]$

We have already discussed that for real roots $\alpha$ & $\beta$, the expression $f(x)=ax^2+bx+c \le 0$ when $x \in [\alpha, \beta]$ for $a>0$. What happens in this case if $f(x)$ does not have any real roots? In which case, the graph never touches the X-Axis – which means $f(x)>0$ for all real $x$. Similarly, for $a<0$ and negative discriminant, the graph of $f(x)$ shall always lie below the X-Axis – thus $f(x)<0$ for all real $x$. It should be very clear that the extremums still lie at $x=-\frac{b}{2a}$ irrespective of the nature of the discriminant.

Ex-1

Find the range of the function $f(x)=\frac{x^2+1}{3x-2}$ for real $x$.

Solution:

Let $\frac{x^2+1}{3x-2} = k$. Thus we have $x^2-3kx+(2k+1)=0$. We must have real roots for this equation – thus the discriminant should be positive.

$9k^2-8k-4 \ge 0$, For this to happen $k$ must not lie in the interval of roots for this equation.

$k \ge \frac{4+2\sqrt{13}}{9}$ & $k \le \frac{4-2\sqrt{13}}{9}$.

This the range of the given function is

$\left(-\infty, \frac{4-2\sqrt{13}}{9}\right] \bigcup \left[\frac{4+2\sqrt{13}}{9}, \infty \right )$

Ex-2

Find all $x$ such that $x^2-x-1 <5$

Solution:

We have $g(x)=x^2-x-1$, for $g(x)<5 \iff x^2-x-6<0 \iff (x-3)(x+2)<0$.

Here the leading coefficient of $f(x)=x^2-x-6$ is 1, thus $f(x)<0$ if and only if $x$ lies between the roots of $f(x)=0$. Thus $x \in (3,-2)$.

We need to note a couple of points over here:

if $f(x_1)g(x_2)<0$, then $f(x_1)$ & $g(x_2)$ are of different signs
if $f(x_1)g(x_2)>0$, then $f(x_1)$ & $g(x_2)$ are of same signs (Both of them may be positive or both of them may be negative)
if $f(x_1)g(x_2)=0$, then at least one of $f(x_1)$ & $g(x_2)$ is of value $0$

Test Your Concepts:

Find all $x$ satisfying $(x-1)(x-2)(1-2x)>0$

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Week -3 | System of Equations – 2

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