Characteristics of Quadratic Functions Section 2.2 beginning on page 56

By the End of This Section … You will be able to identify key aspects of the graph of a function based on its equation in vertex form, intercept form, and standard form. You will understand the significance of the vertex and that the y-value of the vertex is the maximum or minimum of the function while the x-value is when or where that maximum or minimum occurs.

Properties of Parabolas The axis of symmetry is a line that divides a parabola into mirror images. The axis of symmetry passes through the vertex. Vertex form : 𝑓 𝑥 =𝑎 (𝑥−ℎ) 2 +𝑘 The vertex is at the point (ℎ,𝑘) The axis of symmetry is the line 𝑥=ℎ.

First Identify h and k to determine the vertex and axis of symmetry. 𝑥=−3 (−3,4) 𝐴.𝑂.𝑆 𝑥=−3 ℎ=−3 Vertex (−3,4) 𝑘=4 Second, find another point by picking a value of x close to the axis of symmetry and plugging it in to the function to find the y-value that goes with it. 𝑓 −2 =−2 −2+3 2 +4 𝑥=−2 𝑓 −2 =2 (−2,2) Third, reflect that point over the axis of symmetry and draw the parabola through the three points you have plotted. *** I like to plot a fourth and fifth point whenever possible to have a more accurate graph.

Standard Form 𝑓 𝑥 =𝑎 𝑥 2 +𝑏𝑥+𝑐 𝒙= −𝒃 𝟐𝒂 The x-value of the vertex and the axis of symmetry can be found using the formula: The value of 𝒄 in standard form is the y-intercept. (0,𝑐) 𝒙= −𝒃 𝟐𝒂 The y-value of the vertex is found by plugging this x-value into the original equation.

Graphing a Quadratic Function in Standard Form Example 2: Graph 𝑓 𝑥 =3 𝑥 2 −6𝑥+1 Step 1: Identify a, b, and c Step 2: Find the vertex Step 3: Plot the vertex and the axis of symmetry Step 4: Plot the y-intercept and its reflection in the axis of symmetry Step 5: For extra accuracy…Find another point to plot along with its reflection Step 6: Draw a parabola through the points 𝑎=3, 𝑏=−6, 𝑐=1 𝑥= −𝑏 2𝑎 𝑥= 6 2(3) 𝑥=1 𝑦=𝑓 1 =3 (1) 2 −6 1 +1 𝑦=−2 (1,−2) 𝑥=1 𝑐=1 (0,1) 𝑥=3 𝑓 3 =10 (3,10)

Maximum and Minimum Values Because the vertex is the highest or lowest point on a parabola, its y-coordinate is the maximum value (when 𝒂<𝟎) or the minimum value (when 𝒂>𝟎) of the function. The vertex lies on the axis of symmetry so the function is increasing on one side of the axis of symmetry and decreasing on the other side.

Example 3: Find the minimum or maximum value of 𝑓 𝑥 = 1 2 𝑥 2 −2𝑥−1 Example 3: Find the minimum or maximum value of 𝑓 𝑥 = 1 2 𝑥 2 −2𝑥−1. Describe the domain and range of the function and where the function is increasing and decreasing. Is there a maximum or minimum? Find the vertex (the y-value is the max/min) The Domain: The Range: Increasing/Decreasing? 𝑎>0 , there is a minimum 𝑥= −𝑏 2𝑎 𝑥= 2 2(1/2) 𝒙=𝟐 𝑦=𝑓 2 = 1 2 (2) 2 −2 2 −1 𝒚=−𝟑 The minimum is -3 All Real Numbers 𝒚≥−𝟑 Since we have a minimum value, all of the y values will be at or above that minimum value. Because this function has a minimum, it is decreasing to the left of 𝑥=2 (the axis of symmetry) and increasing to the right of 𝑥=2.

Graphing Quadratic Functions Using x-intercepts When the graph of a quadratic function has at least one x-intercept, the function can be written in intercept form, 𝑓 𝑥 =𝑎(𝑥−𝑝)(𝑥−𝑞) where 𝑎≠0.

Step 1: Identify the x-intercepts. Step 2: Find the coordinates of the vertex. Step 3: Draw a parabola through the vertex and the points where the x-intercepts occur. 𝑝=−3 𝑞=1 (−3,0) (1,0) 𝑥= 𝑝+𝑞 2 = −3+1 2 = −2 2 =−1 𝒙=−𝟏 𝑦=𝑓 −1 =−2(−1+3)(−1−1) 𝒚=𝟖 =−2 2 (−2) (−1,8)

Modeling With Mathematics Example 5: The parabola shows the path of your fist golf shot, where x is the horizontal distance (in yards) and y is the corresponding height (in yards). The path of your second shot can be modeled by the function 𝒇 𝒙 =−𝟎.𝟎𝟐𝒙 𝒙−𝟖𝟎 . Which shot travels farther before hitting the ground? Which travels higher? We are comparing the maximum heights and the distance the ball traveled. One shot is represented as a graph, and the other as an equation. The graph shows us that the maximum height is …. The graph shows us that the distance travelled is …. 25 yards The y value of the vertex is the maximum (50,25). 100 yards The difference in the x-values is the distance the ball traveled. (0,0) and (100,0) 𝟏𝟎𝟎−𝟎=𝟎

Modeling With Mathematics Example 5: The parabola shows the path of your fist golf shot, where x is the horizontal distance (in yards) and y is the corresponding height (in yards). The path of your second shot can be modeled by the function 𝒇 𝒙 =−𝟎.𝟎𝟐𝒙 𝒙−𝟖𝟎 . Which shot travels farther before hitting the ground? Which travels higher? To find the max height and distance traveled with the equation we can look at the equation in intercept form. Find the x-intercepts…. Identify the distance travelled… Use the x-intercepts to calculate the maximum height … 𝒇 𝒙 =−𝟎.𝟎𝟐(𝒙−𝟎) 𝒙−𝟖𝟎 Height : 25 yards Distance : 100 yards 𝟎,0 and (𝟖𝟎,0) 80−0=80 Distance traveled = 80 yards Maximum height = 32 yards 𝑥= 𝑝+𝑞 2 = 0+80 2 = 80 2 =45 𝒙=𝟒𝟓 The first shot travels further but the second shot travels higher. 𝑦=𝑓 45 =−0.02(45)(45−80) 𝒚=𝟑𝟐

Finding a Minimum or Maximum

Graphing Quadratic Functions 2) 𝑔 𝑥 =2 (𝑥−2) 2 +5

Graphing a Quadratic Function in Intercept Form