Extreme Values and Graph of a Quadratic Expression
Finding the Minimum/Maximum Value of a Quadratic Expression
To find the minimum or maximum value of a quadratic expression, we use its completed square form.
Starting with the Completed Square Form
Given a quadratic expression \( ax^2 + bx + c \), its completed square form is:
where \( \Delta = b^2  4ac \)

Since \( \left(x + \frac{b}{2a}\right)^2 \) is always nonnegative, the minimum or maximum value of the quadratic expression depends on the sign of \( a \):

If \( a > 0 \), the expression reaches its minimum value when \( \left(x + \frac{b}{2a}\right)^2 = 0 \), that is, \( x = \frac{b}{2a} \).
 The minimum value is \( \frac{\Delta}{4a} \) and it occurs at \( x = \frac{b}{2a} \)

If \( a < 0 \) , the expression reaches its maximum value when \( \left(x + \frac{b}{2a}\right)^2 = 0 \), that is, \( x = \frac{b}{2a} \)
 The maximum value is \( \frac{\Delta}{4a} \) and it occurs at \( x = \frac{b}{2a} \)
The following examples must make this absolutely clear.
Examples
Here are examples demonstrating how to find the minimum and maximum values of given quadratic expressions:
Example 1: Find the Minimum Value
Expression: \( x^2  4x + 3 \)
 Coefficients: \( a = 1 \), \( b = 4 \), \( c = 3 \)
 \( \Delta = b^2  4ac = (4)^2  4 \times 1 \times 3 = 16  12 = 4 \)
 Since \( a > 0 \), the expression has a minimum value.
 Minimum Value: \( \frac{\Delta}{4a} = \frac{4}{4 \times 1} = 1 \)
The minimum value of \( x^2  4x + 3 \) is \( 1 \).
Example 2: Find the Maximum Value
Expression: \( 2x^2 + 4x  1 \)
 Coefficients: \( a = 2 \), \( b = 4 \), \( c = 1 \)
 \( \Delta = b^2  4ac = 4^2  4 \times (2) \times (1) = 16  8 = 8 \)
 Since \( a < 0 \), the expression has a maximum value.
 Maximum Value: \( \frac{\Delta}{4a} = \frac{8}{4 \times (2)} = 1 \)
The maximum value of \( 2x^2 + 4x  1 \) is \( 1 \).
These examples illustrate how to use the formula \( \frac{\Delta}{4a} \) to find the minimum or maximum values of quadratic expressions.
No maximum and No minimum
Consider a quadratic expression of the form \( ax^2 + bx + c \). When \( a > 0 \), its completed square form is:
Let's analyze this:

Squared Term:
 The term \( \left(x + \frac{b}{2a}\right)^2 \) is always nonnegative since it's a square. It equals zero when \( x = \frac{b}{2a} \) and increases as \( x \) moves away from this value.

Coefficient \( a \):
 With \( a > 0 \), multiplying the squared term by \( a \) maintains its nonnegative nature. As \( x \) deviates from \( \frac{b}{2a} \), \( a\left(x + \frac{b}{2a}\right)^2 \) grows larger.

No Upper Bound:
 The value of \( a\left(x + \frac{b}{2a}\right)^2 \) can become arbitrarily large as \( x \) approaches either positive or negative infinity. This indicates no upper limit to the value of the expression.

Constant Term:
 The constant term \( \frac{\Delta}{4a} \) does not influence the unbounded growth of the quadratic expression.
In conclusion, when \( a > 0 \), the quadratic expression in its completed square form clearly demonstrates that as \( x \) moves away from \( \frac{b}{2a} \), the expression's value increases without bound. Thus, there is no maximum value for the expression. With similar reasoning when \(a<0\), the quadratic expression has no minimum value.
Very Important
The formula to calculate maximum and minimum values is based on the assumption that the input is allowed to change for all possible real values, that is, \(x\in\mathbb R\). But this is not always so. The input might be restricted a subset of \(\mathbb R\). In that case, we adhere to the ideas discussed below.
Drawing Graph of a quadratic Expression
To learn how to draw the graph of a quadratic expression, we start with the simplest possible quadratic expression, that is, \(x^2\).
To draw the graph of the quadratic expression \( x^2 \), follow these steps:

Choose a Range of xvalues: Select a range of xvalues to plot. For symmetry, include both positive and negative values. For example, choose values like \(5, 4, 3, \ldots, 3, 4, 5\).

Calculate Corresponding yvalues: For each xvalue, calculate the yvalue, which is \( x^2 \).

Create a Table of Points: Organize the xvalues and their corresponding yvalues in a table:
x y 5 25 4 16 3 9 2 4 1 1 0 0 1 1 2 4 3 9 4 16 5 25 
Plot the Points on a Graph: On a coordinate plane, plot each of the points from your table.

Draw the Curve: Connect these points with a smooth curve. The graph of \( x^2 \) is called a parabola that opens upwards.
Parabola
 The graph of \(x^2\) is called parabola.
 The point \((0,0)\) is called the vertex of this parabola. Observe that it is also the point where the value of \(x^2\) is minimum.
 The graph is symmetrical around yaxis. The line about which the parabola is symmetrical is called the axis or axis of symmetry of the parabola.
Obtaining Graphs via Transformation
The graph of any quadratic expression can be drawn by transforming \(x^2\). To understand that first we learn about simple transformations, which can be combined together to achieve our goal.
I. Horizontal Shifting
Draw the graph of \((x1)^2\) by transforming the graph of \(x^2\).
With this exercise we will learn about the horizontal shift.

Start with the Graph of \(x^2\):
 The graph of \(x^2\) is a parabola that opens upwards with its vertex at the origin (0,0).

Understand the Expression \((x1)^2\):
 This expression can be interpreted as the original \(x^2\) function with a modification: replacing \(x\) with \(x1\).
 This replacement means that every xcoordinate in the original graph will be shifted.

Horizontal Shift Explanation:
 The term \((x1)\) indicates a horizontal shift.
 Algebraically, substituting \(x\) with \(x1\) implies that what was at \(x\) in the original graph now occurs at \(x1\).
 For example, the vertex of \(x^2\) at (0,0) shifts to where \(x\) is \(1\) in \((x1)^2\).

Shift to the Right:
 The substitution \(x \rightarrow x1\) translates to moving each point on the graph of \(x^2\) one unit to the right.
 This results in the entire graph shifting one unit to the right.

Drawing the New Graph:
 To draw \((x1)^2\), shift every point of the \(x^2\) parabola one unit to the right.
 The new vertex will be located at (1,0), with the parabola still opening upwards.
In summary, the graph of \((x1)^2\) is obtained by horizontally shifting the graph of \(x^2\) one unit to the right, due to the substitution \(x \rightarrow x1\) in the function.
Generalization of Horizontal Shifting
To generalize the transformation for the quadratic expression \((x  a)^2\), where \(a\) is a real number, consider two cases  forward and backward shifts:

Forward Shift (when \(a > 0\)):
 The expression \((x  a)^2\) shifts the graph of \(x^2\) to the right by \(a\) units.
 The vertex of the parabola, originally at \((0,0)\), moves to \((a,0)\).

Backward Shift (when \(a < 0\)):
 For negative \(a\), \((x  a)^2\) shifts the graph of \(x^2\) to the left. Since \(a\) is negative, shifting left by \(a\) units means the vertex is at \((a,0)\).
 The vertex moves from \((0,0)\) to \((a,0)\), reflecting the negative value of \(a\).
In both cases, the shape of the parabola (upward opening) remains unchanged, with the position of the vertex along the xaxis being the primary change.
II. Vertical Shifting
Let us uderstand with example what is vertical shifting by drawing the graph of \(y=x^23\).
Here's the process:

Start with the Graph of \( y = x^2 \):
 The graph of \( y = x^2 \) is a parabola that opens upwards, with its vertex at the origin (0,0).

Apply a Vertical Shift:
 The expression \( y = x^2  3 \) indicates a vertical shift of the graph of \( y = x^2 \).
 The "3" in the equation means you lower the entire graph of \( y = x^2 \) down by 3 units.

Shift the Vertex:
 The original vertex of \( y = x^2 \) at (0,0) shifts down by 3 units to (0, 3).

Resulting Graph:
 The entire graph shifts vertically downwards by 3 units.
 The new graph represents \( y = x^2  3 \), with its vertex now at (0, 3).

Draw the New Graph:
 To draw \( y = x^2  3 \), shift every point of the \( y = x^2 \) parabola down by 3 units.
 The shape of the parabola remains unchanged, still opening upwards, but its vertex is now at (0, 3).
In summary, to draw the graph of \( y = x^2  3 \), vertically shift the graph of \( y = x^2 \) downwards by 3 units. The vertex of the parabola moves from (0,0) to (0, 3).
Generalization of Vertical Shifting
To generalize the vertical shifting of the quadratic expression \( y = x^2 + c \), where \( c \) is a real number, consider two cases  upward and downward shifts:

Upward Shift (when \( c > 0 \)):
 The expression \( y = x^2 + c \) represents a vertical shift of the graph of \( y = x^2 \) upwards by \( c \) units.
 The vertex of the parabola, originally at (0,0), moves to (0, c).

Downward Shift (when \( c < 0 \)):
 The expression \( y = x^2 + c \) for negative \( c \) shifts the graph of \( y = x^2 \) downwards by the absolute value of \( c \) units.
 The vertex of the parabola shifts from (0,0) to (0, c), reflecting the negative value of \( c \).
In both cases, the shape of the parabola (upward opening) remains unchanged, with the position of the vertex along the yaxis being the primary change.
III. Combining Horizontal and Vertical Shifting
We can combine the horizontal and vertical shifting to draw the graph of more complex quadratic expressions. For instance, we draw the graph of \(y=(x+1)^23\).
To draw the graph of \( y = (x + 2)^2  3 \) by combining horizontal and vertical shifting, follow these steps:

Start with the Graph of \( y = x^2 \):
 The graph of \( y = x^2 \) is a parabola opening upwards, with its vertex at the origin (0,0).

Apply Horizontal Shift:
 The expression \( (x + 2)^2 \) indicates a horizontal shift of the graph of \( y = x^2 \) to the left by 2 units.
 This moves the vertex of the parabola from (0,0) to (2,0).

Apply Vertical Shift:
 The "3" in the expression \( (x + 2)^2  3 \) indicates a vertical shift downwards by 3 units.
 After the horizontal shift, the vertex at (2,0) is moved down to (2, 3).

Combine the Shifts:
 First, shift the graph of \( y = x^2 \) horizontally to the left by 2 units.
 Then shift this graph downwards by 3 units.

Resulting Graph:
 The final graph of \( y = (x + 2)^2  3 \) is a parabola that opens upwards, with its vertex located at (2, 3).
In summary, to draw \( y = (x + 2)^2  3 \), combine a 2unit horizontal shift to the left and a 3unit vertical shift downwards. The vertex of the parabola originally at (0,0) moves to (2, 3) due to these shifts.
IV. Vertical Streching and Compression along yaxis
We can understand this by comparing the graphs of \( y = \frac{x^2}{3} \) and \( y = 3x^2 \) and examining how the coefficients affect the shape of the parabola relative to the standard parabola \( y = x^2 \):

Graph of \( y = 3x^2 \):
 The coefficient '3' in \( y = 3x^2 \) multiplies the yvalues of the standard parabola \( y = x^2 \) by 3.
 This results in a vertical stretch, making the parabola steeper and narrower compared to \( y = x^2 \).
 For any given xvalue, the height (yvalue) on this graph is three times that of the standard parabola.

Graph of \( y = \frac{x^2}{3} \):
 In \( y = \frac{x^2}{3} \), the coefficient \( \frac{1}{3} \) reduces each yvalue of \( y = x^2 \) to onethird.
 This leads to a vertical compression, making the parabola wider and less steep than \( y = x^2 \).
 The height of the parabola for any xvalue is only onethird compared to \( y = x^2 \).
In summary:
 \( y = 3x^2 \) represents a vertically stretched parabola, resulting in a steeper and narrower shape.
 \( y = \frac{x^2}{3} \) represents a vertically compressed parabola, resulting in a wider and less steep shape.
 Both transformations alter the steepness and width of the parabola while maintaining its basic shape and symmetry.
Generalization of Vertical Stretching and Compression
The graph of the quadratic expression \( y = ax^2 \), where \( a > 0 \), can be understood as a transformation of the standard parabola \( y = x^2 \) through vertical scaling:

Effect of the Coefficient \( a \) on the Parabola:
 The coefficient \( a \) in \( y = ax^2 \) determines the degree of vertical stretching or compression of the parabola.
 Since \( a > 0 \), the parabola always opens upwards.

Vertical Stretching (when \( a > 1 \)):
 If \( a \) is greater than 1, the graph of \( y = ax^2 \) is a vertically stretched version of \( y = x^2 \).
 The parabola becomes steeper and narrower compared to \( y = x^2 \).
 The vertex remains at the origin (0,0), but the points on the parabola are farther from the xaxis for any given xvalue.

Vertical Compression (when \( 0 < a < 1 \)):
 If \( a \) is between 0 and 1, the graph of \( y = ax^2 \) is a vertically compressed version of \( y = x^2 \).
 The parabola becomes wider and less steep.
 The points on the parabola are closer to the xaxis for the same xvalues compared to \( y = x^2 \).
In summary, for \( y = ax^2 \) with \( a > 0 \), the graph is a vertically scaled version of \( y = x^2 \). The value of \( a \) determines whether the parabola is stretched or compressed vertically, but it always retains its upwardopening shape.
V. Reflection across xaxis
Reflection across the xaxis is a transformation that inverts the graph of a function over the xaxis. This happens when we multiply expression \(x^2\) with \(1\).
Properties of Reflection Across the XAxis:

Inversion of Points:
 Points that are above the xaxis on the original graph are reflected to below the xaxis on the transformed graph, and vice versa.
 If a point on the original graph is at \((x, y)\), it will be at \((x, y)\) on the reflected graph.

Effect on Graph Shape:
 The overall shape of the graph is preserved, but it is flipped over the xaxis.
 For example, an upwardopening parabola \( y = x^2 \) becomes a downwardopening parabola \( y = x^2 \) after reflection.
VI. Combining all transformations
Let us learn how to draw the graph of \( y = 2x^2 + 4x  1 \) starting from the graph of \( y = x^2 \). Follow these steps:

Convert to Completed Square Form: Start by converting \( y = 2x^2 + 4x  1 \) into its completed square form:
 Factor out the coefficient of \( x^2 \): \( y = 2(x^2 + 2x)  1 \).

Complete the square:
\[ y = 2\left(x^2 + 2x + 1  1\right)  1 \]\[ y = 2\left((x + 1)^2  1\right)  1 \] 
Simplify:
\[ y = 2(x + 1)^2  3 \]

Start with the Graph of \( y = x^2 \):
 The basic parabola \( y = x^2 \) opens upwards with its vertex at (0,0).

Apply Transformations:
 Horizontal Shift: Shift the graph to the left by 1 unit due to \((x + 1)\), moving the vertex to (1,0).
 Vertical Stretch: Stretch the graph vertically by a factor of 2, making the parabola narrower.
 Vertical Shift: Shift the graph downwards by 3 units because of the \(3\), moving the vertex to (1,3).

Resulting Graph:
 The graph of \( y = 2(x + 1)^2  3 \) is a vertically stretched and narrower parabola, shifted left by 1 unit and down by 3 units, with its vertex at (1, 3).
In summary, to graph \( y = 2x^2 + 4x  1 \), start with \( y = x^2 \), apply a horizontal shift left, a vertical stretch, and a vertical shift downwards, transforming the vertex to (1, 3).
Generalization
To graph the quadratic expression \( y = ax^2 + bx + c \) with a focus on the case when \( a < 0 \), follow these steps:

Completed Square Form:

Convert \( y = ax^2 + bx + c \) to its completed square form:
\[ y = a\left(x + \frac{b}{2a}\right)^2  \frac{\Delta}{4a} \]where \( \Delta = b^2  4ac \).

The vertex of the parabola is at \( \left(\frac{b}{2a}, \frac{\Delta}{4a}\right) \), as we show below.


Start with \( y = x^2 \):
 Begin with the basic graph of \( y = x^2 \), a parabola opening upwards with its vertex at (0,0).

Apply Transformations:
 Horizontal Shift: Shift the graph horizontally to \( x = \frac{b}{2a} \). The vertex moves to \( \left(\frac{b}{2a}, 0\right) \).
 Vertical Stretch/Compression and Reflection:
 If \( a > 1 \) or \( 0 < a < 1 \), stretch or compress the graph vertically. The vertex remains at \( \left(\frac{b}{2a}, 0\right) \).
 If \( a < 0 \), in addition to stretching or compressing, the graph is reflected across the xaxis, changing the direction of the parabola to open downwards.
 Vertical Shift: Shift the graph vertically by \( \frac{\Delta}{4a} \). The vertex moves to its final position at \( \left(\frac{b}{2a}, \frac{\Delta}{4a}\right) \).
By following these steps and applying the transformations, you can graph any quadratic expression \( y = ax^2 + bx + c \). The vertex location \( \left(\frac{b}{2a}, \frac{\Delta}{4a}\right) \) is crucial, especially when \( a < 0 \), as it indicates a downwardopening parabola due to the reflection across the xaxis.
Drawing Graph by locating vertex
Note
What we discuss now shortens the process of drawing graphs and should be used as a standard way of drawing graphs of quadratic expressions. Nonetheless the above process was important to understand, as it gives you first introduction to transformation process of drawing graphs. You shall see this again in functions.
Because of the symmetry of the graph of quadratic expression, and its definite shape, we can draw the graph by knowing vertex of the parabola and atleast one more point on the graph (generally, the yintercept of the graph, that is, where \(x=0\)). Here's how we draw the graph:

Find the Vertex:
 Using the Vertex Formula: The vertex is at \( \left(\frac{b}{2a}, \frac{\Delta}{4a}\right) \), where \( \Delta = b^2  4ac \). This formula gives you the coordinates of the vertex directly.
 Using Substitution: First, find the xcoordinate of the vertex with \( x = \frac{b}{2a} \). Then, substitute this xvalue back into the original equation to find the ycoordinate: \( y = a\left(\frac{b}{2a}\right)^2 + b\left(\frac{b}{2a}\right) + c \).

Plot the Vertex:
 Plot the vertex on the graph using either of the methods above to find its coordinates.

Find the YIntercept:
 The yintercept is the value of \( y \) when \( x = 0 \), which is simply \( y = c \). Plot this point as (0, c) on the graph.

Effect of the Coefficient \( a \):
 If \( a > 0 \), the parabola opens upwards.
 If \( a < 0 \), it opens downwards.
 The value of \( a \) determines the width of the parabola; larger \( a \) values create narrower parabolas.

Draw the Parabola:
 Begin from the vertex, ensuring the parabola opens in the correct direction based on the sign of \( a \).
 The parabola should pass through the yintercept and be symmetrical around the vertex.
Using the above process, you can accurately plot the graph of any quadratic expression \( y = ax^2 + bx + c \), taking into account the shape and position influenced by the coefficients \( a \), \( b \), and \( c \).
Let us apply this on the following example:
Let us draw the graph of the quadratic expression \( y = 3x^2 + 4x + 9 \), using the following steps:

Find the Vertex:

Using the Vertex Formula: The vertex can be found using the formula \( \left(\frac{b}{2a}, \frac{\Delta}{4a}\right) \), where \( \Delta = b^2  4ac \). For our equation:
\[ a = 3, \quad b = 4, \quad c = 9 \]\[ \Delta = 4^2  4 \cdot (3) \cdot 9 = 16 + 108 = 124 \]\[ \text{Vertex} = \left(\frac{4}{2 \cdot (3)}, \frac{124}{4 \cdot (3)}\right) = \left(\frac{2}{3}, \frac{31}{6}\right) \]


Plot the Vertex:
 Plot the vertex at \( \left(\frac{2}{3}, \frac{31}{6}\right) \) on the graph.

Find the YIntercept:\
 The yintercept is \( y = c \), which in this case is \( y = 9 \). Plot the point (0, 9).

Effect of the Coefficient \( a \):
 Since \( a = 3 \) (and \( a < 0 \)), the parabola opens downwards and is vertically stretched, making it narrower.

Draw the Parabola:
 Starting from the vertex, draw a narrow, downwardopening parabola passing through the yintercept at (0, 9).
By following these steps, the graph of \( y = 3x^2 + 4x + 9 \) can be accurately plotted, taking into account its vertex, yintercept, and the shape influenced by the coefficient \( 3 \).