Quadratic Functions and Their Graphs Section 10.4 Quadratic Functions and Their Graphs

Objectives Graph functions of the form ƒ(x) = ax2 and ƒ(x) = ax2 + k Graph functions of the form ƒ(x) = a(x – h)2 and ƒ(x) = a(x – h)2 + k Graph functions of the form ƒ(x) = ax2 + bx + c by completing the square Find the vertex using Determine minimum and maximum values Solve quadratic equations graphically

Objective 1: Graph Functions of the Form ƒ(x) = ax2 and ƒ(x) = ax2 + k A quadratic function is a second-degree polynomial function that can be written in the form ƒ(x) = ax2 + bx + c, where a, b, and c are real numbers and a ≠ 0. Quadratic functions are often written in another form, called standard form, ƒ(x) = a(x – h)2 + k, where a, h, and k are real numbers and a ≠ 0. This form is useful because a, h, and k give us important information about the graph of the function. To develop a strategy for graphing quadratic functions written in standard form, we will begin by considering the simplest case, ƒ(x) = ax2. One way to graph quadratic functions is to plot points.

Objective 1: Graph Functions of the Form ƒ(x) = ax2 and ƒ(x) = ax2 + k The graphs of functions of the form ƒ(x) = ax2 are parabolas. The lowest point on a parabola that opens upward, or the highest point on a parabola that opens downward, is called the vertex of the parabola. The vertical line, called an axis of symmetry, that passes through the vertex divides the parabola into two congruent halves. If we fold the paper along the axis of symmetry, the two sides of the parabola will match.

Objective 1: Graph Functions of the Form ƒ(x) = ax2 and ƒ(x) = ax2 + k The graph of ƒ(x) = ax2 is a parabola opening upward when a > 0 and downward when a < 0, with vertex at the point (0, 0) and axis of symmetry the line x = 0.

EXAMPLE 1 Graph: Strategy We can make a table of values for each function, plot each point, and connect them with a smooth curve. Why At this time, this method is our only option.

EXAMPLE 1 Solution Graph: After graphing each curve, we see that the graph of g(x) = 3x2 is narrower than the graph of ƒ(x) = x2, and the graph of is wider than the graph of ƒ(x) = x2. For ƒ (x) = ax2, the smaller the value of | a |, the wider the graph.

EXAMPLE 1 Graph: Solution

Objective 1: Graph Functions of the Form ƒ(x) = ax2 and ƒ(x) = ax2 + k The graph of ƒ(x) = ax2 + k is a parabola having the same shape as ƒ(x) = ax2 but translated k units upward if k is positive and | k | units downward if k is negative. The vertex is at the point (0, k), and the axis of symmetry is the line x = 0.

EXAMPLE 3 Graph: Strategy We make a table of values for each function, plot each point, and connect them with a smooth curve. Why At this time, this method is our only option.

EXAMPLE 3 Graph: Solution

EXAMPLE 3 Solution Graph: After graphing the curves, we see that the graph of g(x) = 2x2 + 3 is identical to the graph of ƒ(x) = 2x2, except that it has been translated 3 units upward. The graph of s (x) = 2x2 – 3 is identical to the graph of ƒ(x) = 2x2 , except that it has been translated 3 units downward. In each case, the axis of symmetry is the line x = 0.

Objective 2: Graph Functions of the Form ƒ(x) = a(x – h)2 and ƒ(x) = a(x – h)2 + k The graph of ƒ(x) = a(x – h)2 is a parabola having the same shape as ƒ(x) = ax2 but translated h units to the right if h is positive and | h | units to the left if h is negative. The vertex is at the point (h, 0), and the axis of symmetry is the line x = h.

EXAMPLE 4 Graph: Strategy We make a table of values for each function, plot each point, and connect them with a smooth curve. Why At this time, this method is our only option.

EXAMPLE 4 Graph: Solution

EXAMPLE 4 Solution Graph: We note that the graph of g(x) = 2(x – 3)2 is identical to the graph of ƒ(x) = 2x2 , except that it has been translated 3 units to the right. The graph of s(x) = 2(x + 3)2 is identical to the graph of ƒ(x) = 2x2 , except that it has been translated 3 units to the left.

Objective 2: Graph Functions of the Form ƒ(x) = a(x – h)2 and ƒ(x) = a(x – h)2 + k The graph of the quadratic function ƒ(x) = a(x – h)2 + k where a ≠ 0, is a parabola with vertex at (h, k). The axis of symmetry is the line x = h. The parabola opens upward when a > 0 and downward when a < 0.

EXAMPLE 5 Graph: ƒ(x) = 2(x – 3)2 – 4. Label the vertex and draw the axis of symmetry. Strategy We will determine whether the graph opens upward or downward and find its vertex and axis of symmetry. Then we will plot some points and complete the graph. Why This method will be more efficient than plotting many points.

EXAMPLE 5 Graph: ƒ(x) = 2(x – 3)2 – 4. Label the vertex and draw the axis of symmetry. Solution The graph of ƒ(x) = 2(x – 3)2 – 4 is identical to the graph of g(x) = 2(x – 3)2, except that it has been translated 4 units downward. The graph of g(x) = 2(x – 3)2 is identical to the graph of s(x) = 2x2, except that it has been translated 3 units to the right.

EXAMPLE 5 Graph: ƒ(x) = 2(x – 3)2 – 4. Label the vertex and draw the axis of symmetry. Solution

Objective 3: Graph functions of the form ƒ(x) = ax2 + bx + c by completing the square. To graph functions of the form ƒ(x) = ax2 + bx + c, we can complete the square to write the function in standard form ƒ(x) = a(x – h)2 + k.

EXAMPLE 6 Determine the vertex and the axis of symmetry of the graph of ƒ(x) = x2 + 8x + 21. Will the graph open upward or downward? Strategy To find the vertex and the axis of symmetry, we will complete the square on x and write the equation of the function in standard form. Why Once the equation is written in standard form, we can determine the values of a, h, and k. The coordinates of the vertex will be (h, k) and the equation of the axis of symmetry will be x = h. The graph will open upward if a > 0 or downward if a < 0.

EXAMPLE 6 Determine the vertex and the axis of symmetry of the graph of ƒ(x) = x2 + 8x + 21. Will the graph open upward or downward? Solution

EXAMPLE 6 Determine the vertex and the axis of symmetry of the graph of ƒ(x) = x2 + 8x + 21. Will the graph open upward or downward? Solution

Objective 4: Find the Vertex using Because of symmetry, if a parabola has two x-intercepts, the x-coordinate of the vertex is exactly midway between them. We can use this fact to derive a formula to find the vertex of a parabola. Formula for the Vertex of a Parabola: The vertex of the graph of the quadratic function ƒ(x) = ax2 + bx + c is and the axis of symmetry of the parabola is the line

Objective 4: Find the Vertex using Graphing a Quadratic Equation ƒ(x) = ax2 + bx + c. Determine whether the parabola opens upward or downward by finding the value of a. The x-coordinate of the vertex of the parabola is . To find the y-coordinate of the vertex, substitute for x and find . The axis of symmetry is the vertical line passing through the vertex. The y-intercept is determined by the value of ƒ(x) when x = 0: the y-intercept is (0, c) . The x-intercepts (if any) are determined by the values of x that make ƒ(x) = 0. To find them, solve the quadratic equation ax2 + bx + c = 0.

EXAMPLE 8 Determine the vertex of the graph of ƒ(x) = 2x2 – 4x – 1. Strategy We will determine the values of a and b and substitute into the formula for the vertex of a parabola. Why It is easier to find the coordinates of the vertex using the formula than it is to complete the square on 2x2 – 4x – 1.

EXAMPLE 8 Determine the vertex of the graph of ƒ(x) = 2x2 – 4x – 1. Solution The function is written in ƒ(x) = ax2 + bx + c form, where a = 2 and b = –4. To find the vertex of its graph, we compute The vertex is the point (1, –3).

Objective 5: Determine Minimum and Maximum Values It is often useful to know the smallest or largest possible value a quantity can assume. For example, companies try to minimize their costs and maximize their profits. If the quantity is expressed by a quadratic function, the y-coordinate of the vertex of the graph of the function gives its minimum or maximum value.

EXAMPLE 10 Minimizing Costs. A glassworks that makes lead crystal vases has daily production costs given by the function C(x) = 0.2x2 – 10x + 650, where x is the number of vases made each day. How many vases should be produced to minimize the per-day costs? What will the costs be? Strategy We will find the vertex of the graph of the quadratic function. Why The x-coordinate of the vertex indicates the number of vases to make to keep costs at a minimum, and the y-coordinate indicates the minimum cost.

EXAMPLE 10 Minimizing Costs. A glassworks that makes lead crystal vases has daily production costs given by the function C(x) = 0.2x2 – 10x + 650, where x is the number of vases made each day. How many vases should be produced to minimize the per-day costs? What will the costs be? Solution The graph of C(x) = 0.2x2 – 10x + 650 is a parabola opening upward. The vertex is the lowest point on the graph. To find the vertex, we calculate The vertex is (25, 525), and it indicates that the costs are a minimum of $525 when 25 vases are made daily.

Objective 6: Solve Quadratic Equations Graphically When solving quadratic equations graphically, we must consider three possibilities. If the graph of the associated quadratic function has two x-intercepts, the quadratic equation has two real-number solutions. Figure (a) shows an example of this. If the graph has one x-intercept, as shown in figure (b), the equation has one repeated real-number solution. Finally, if the graph does not have an x-intercept, as shown in figure (c), the equation does not have any real-number solutions.