Unit 3, Lesson 3.1 Creating and Graphing Equations Using Standard Form

Transcription

1 Unit 3, Lesson 3.1 Creating and Graphing Equations Using Standard Form Imagine the path of a basketball as it leaves a player s hand and swooshes through the net. Or, imagine the path of an Olympic diver as she arcs away from the diving board, slips deep into the water, and resurfaces. Both traveling paths make a U-shaped curve known as a parabola. A parabola is the U-shaped graph of a quadratic function, which is an equation with a degree of. Formally, a quadratic function is a function that can be written in the form f(x) = ax + bx + c, where a 0. Quadratics can be used to model varying situations. In this lesson, you will learn about and practice identifying the important points on a parabola and use those points to sketch the corresponding graph. The standard form of a quadratic function is f(x) = ax + bx + c, where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant term. The x-intercepts of the graph of a quadratic function are the points at which the graph crosses the x-axis, and are written as (x, 0). The y-intercept of the graph of a quadratic function is the point at which the graph crosses the y-axis and is written as (0, y). The graph below shows the location of the parabola s y-intercept. The equation of the parabola is y = x x 4. Note that the y-intercept of this equation is 4 and the constant term of the quadratic equation is 4. The y-intercept of a quadratic is the c value of the quadratic equation when written in standard form. The axis of symmetry of a parabola is the line through the vertex of a parabola about which the parabola is symmetric. MUnit3Lesson /19/018

2 The axis of symmetry extends through the vertex of the graph. The vertex of a parabola is the point on a parabola where the graph changes direction, (h, k), where h is the x-coordinate and k is the y-coordinate. The equation of the axis of symmetry is. a In the above example, the equation of the axis of symmetry is 1 because the vertical line through 1 is the line that cuts the parabola in half. The vertex of a quadratic lies on the axis of symmetry, as shown in the graph below. The formula is not only used to find the equation of the axis of symmetry, but also to find the a x-coordinate of the vertex. To find the y-coordinate, substitute the value of x into the original function, ( h,k ), f. a a f. Thus, a The vertex gives information about the maximum or minimum value of a quadratic. The maximum is the largest y-value of a quadratic and the minimum is the smallest y-value. One might see these terms in real-world situations such as maximizing profit and minimizing cost, among others. From the equation of a function in standard form, you can determine if you have a maximum or minimum based on the sign of the coefficient of the quadratic term, a. If a > 0, the parabola opens up and therefore has a minimum value. If a < 0, the parabola opens down and therefore has a maximum value. MUnit3Lesson3.1 8/19/018

3 Minimum Maximum y 3x y 3x a 3 and a 0 a 3 and a 0 To create the equation of a quadratic in standard form, one must know some combination of the following key features: y-intercept, axis of symmetry or the vertex, and maximum or minimum. The key features of a quadratic are the x-intercepts, y-intercept, where the function is increasing and decreasing, where the function is positive and negative, relative minimums and maximums, symmetries, and end behavior of the function. To create the equation of a quadratic given the vertex and y-intercept, first set the x-coordinate of the vertex equal to and isolate b. a Second, substitute the expression that is equal to b, the y-intercept, and the coordinates of the vertex into the standard form of a quadratic equation, y = ax + bx + c. Solve the equation for a and then substitute the value of a into the equation you created in the first step to determine the value of b. Common Errors/Misconceptions forgetting to write an equation for the axis of symmetry not finding and labeling all important points drawing the parabola as a straight line or a series of straight lines instead of a curve Example 1 Find the y-intercept and vertex of the function f ( x ) x 4 3. Determine whether the vertex is a minimum or maximum point on the graph. 1. Determine the y-intercept. The function f ( x ) x 4 3 is written in standard form, f ( x ) a b c. When 0, the y-intercept equals c, which is 3. Verify that 3 is the y-intercept. MUnit3Lesson /19/018

4 Example 1 (continued) f ( x ) f (0) f (0) f (0) x 4 3 Original equation (0) 4(0) 3 Substitute 0 for x Simplify 3 The y-intercept is the point ( 0,3 ).. Determine the vertex of the function by using ( h,k ), f, where h is the a a x-coordinate of the vertex and k is the function, f ( x ), evaluated for h. f ( x ) x 4 3 is in standard form; therefore, a and b 4. a Equation to determine the x-coordinate of the vertex 4 ( ) Substitute for a and 4 for b 4 Simplify 4 1 The vertex has an x-coordinate of 1. Since h is the x-coordinate of the vertex, or the input value of the vertex, you can find the output value, k, by evaluating the function for f (1). f ( x ) f (1 ) f (1 ) f (1 ) x 4 3 Original equation (1 ) 4(1) 3 Substitute 1 for x (1 ) 4 3 Simplify 7 f (1 ) 5 The y-coordinate of the vertex is 5. The vertex is the point ( 1,5 ). 3. Use the value of a to determine if the vertex is a maximum of a minimum value. Example Given that f ( x ) x 4 3 is written in standard form, a. Since a < 0, the parabola opens downward. As such, the vertex ( 1,5 ) is a maximum point. h( x ) 11x 5 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x-intercept(s), and the y-intercept. Use this information to sketch the graph. 1. Determine whether the graph opens up or down. h( x ) 11x 5 is written in standard form; therefore, a. Since a > 0, the parabola opens upward. MUnit3Lesson /19/018

5 Example (continued). Find the vertex and the equation of the axis of symmetry. h( x ) 11x 5 is in standard form; therefore, a and b 11. a Equation to determine the x-coordinate of the vertex ( 11) ( ) Substitute for a and 11 for b 11 Simplify 4 The vertex has an x-coordinate of Since the input value is, find the output value by evaluating the function for 4 h. 4 h( x ) 11x 5 Original equation h h h h h 4 8 The y-value of the vertex is Substitute 4 11 for x Simplify The vertex is the point, or (.75, ). 4 8 Since the axis of symmetry is the vertical line through the vertex, the equation of the axis 11 of symmetry is or Find the y-intercept. To find the y-intercept, set 0. h(0) (0) 11(0) 5 h( 0 ) h( 0 ) 5 The y-intercept is ( 0,5 ). MUnit3Lesson /19/018

6 Example (continued) 4. Find the x-intercept(s), if any exist. To find the x-intercept, set h( x ) 0 and solve for x. x This equation is factorable, but if you cannot easily identify the factors, the quadratic formula will always work. x ( 1)( 5 ) 0 ( 1) 0 or ( 5 ) 0 x The x-intercepts are,0 and ( 5,0 ). 5. Plot the points from Steps -4 and their symmetric points over the axis of symmetry. Connect the points with a smooth curve. Example 3 x ) 8 8 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x-intercept(s), and the y-intercept. Use this information to sketch the graph. 1. Determine whether the graph opens up or down. x ) 8 8 is written in standard form; therefore, a. Since a > 0, the parabola opens upward. MUnit3Lesson /19/018

7 Example 3 (continued). Find the vertex and the equation of the axis of symmetry. x ) 8 8 is in standard form; therefore, a and b 8. a Equation to determine the x-coordinate of the vertex ( 8 ) ( ) Substitute for a and 8 for b 8 Simplify 4 The vertex has an x-coordinate of. Since the input value is, find the output value by evaluating the function for ). x ) 8 8 Original equation ) ( ) 8( ) 8 Substitute for x ) ( 4 ) 16 8 Simplify ) 8 8 ) 0 The y-coordinate of the vertex is 0. The vertex is the point (,0 ). Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is. 3. Find the y-intercept. To find the y-intercept, set 0. 0) (0) 8(0) 8 0 ) ( 0 ) 0 8 R ( 0 ) ) 8 The y-intercept is ( 0,8 ). 4. Find the x-intercept(s), if any exist. To find the x-intercept, set x ) 0 and solve for x. x ) 0 x This equation is factorable, but if you cannot easily identify the factors, the quadratic formula will always work. x ( 4 4 ) 0 ( )( ) 0 ( ) 0, a double root The x-intercept is (,0 ). This is a special case where the vertex is also the x-intercept. MUnit3Lesson /19/018

8 Example 3 (continued) 5. Plot the points from Steps -4 and their symmetric points over the axis of symmetry. For a more accurate graph, determine an additional pair of symmetric points. Choose any x-coordinate on the left or right of the axis of symmetry. Evaluate the function for the chosen value of x to determine the output value. Let s choose 1. x ) 8 8 Original equation 1) ( 1) 8( 1) 8 Substitute 1 for x 1) (1) 8 8 Simplify 1) 0 1) So, ( 1, ) is an additional point. Plot ( 1, ) on the same graph. ( 1, ) is one unit away from the axis of symmetry. Locate the point that is symmetric to the point ( 1, ). ( 3, ) is also 1 unit away from the axis of symmetry and is symmetrical to the original point ( 1, ). You can verify that ( 0,8 ) and ( 4,8 ) are the same distance from the axis of symmetry and are also symmetrical by referring to the graph. MUnit3Lesson /19/018

9 Example 4 g( x ) 8 17 is a quadratic function. Determine the direction in which the function opens, the vertex, the axis of symmetry, the x-intercept(s), and the y-intercept. Use this information to sketch the graph. 1. Determine whether the graph opens up or down. g( x ) 8 17 is written in standard form; therefore, a 1. Since a < 0, the parabola opens downward.. Find the vertex and the equation of the axis of symmetry. g( x ) 8 17 is in standard form; therefore, a 1 and b 8. a Equation to determine the x-coordinate of the vertex (8) ( 1) Substitute 1 for a and 8 for b 8 Simplify 4 The vertex has an x-coordinate of 4. Since the input value is 4, find the output value by evaluating the function for g ( 4 ). g( x ) 8 17 Original equation g( 4 ) ( 4 ) 8( 4 ) 17 Substitute 4 for x g( 4 ) (16 ) 3 17 Simplify g( 4 ) g( 4 ) 1 The y-coordinate of the vertex is 1. The vertex is the point ( 4, 1). Since the axis of symmetry is the vertical line through the vertex, the equation of the axis of symmetry is Find the y-intercept. To find the y-intercept, set 0. g(0) (0) 8(0) 17 g( 0 ) g( 0 ) 0 17 g( 0 ) 17 The y-intercept is ( 0, 17 ). 4. Find the x-intercept(s), if any exist. To find the x-intercept, set g( x ) 0 and solve for x. g( x ) 0 x This equation is not factorable, so solve using the quadratic formula. MUnit3Lesson /19/018

10 Example 4 (continued) x Determine the values of a, b, and c. a 1, b 8, and c 17 b b 4ac a Quadratic formula 8 8 4( 1)( 17 ) ( 1) Substitute 1 for a, 8 for b, and 17 for c ( 17 ) Since the discriminant is negative, there are no real solutions. This also means there are no x-intercepts. 5. Plot the points from Steps -4 and their symmetric points over the axis of symmetry. For a more accurate graph, determine an additional pair of symmetric points. Choose any x-coordinate on the left or right of the axis of symmetry. Evaluate the function for the chosen value of x to determine the output value. Let s choose 1. g( x ) 8 17 Original equation g(1) (1) 8(1) 17 Substitute 1 for x g(1) (1) 8 17 Simplify g(1) 1 9 g(1) 10 MUnit3Lesson /19/018

11 Example 4 (continued) ( 1, 10) is an additional point. Plot ( 1, 10) on the same graph. ( 1, 10) is 3 units from the axis of symmetry. Locate the point that is symmetric to the point ( 1, 10). ( 7, 10) is also 3 units from the axis of symmetry and is symmetrical to the original point ( 1, 10). You can verify that ( 1, 10) and ( 7, 10) are the same distance from the axis of symmetry and are also symmetrical by referring to the graph. Example 5 Create the equation of a quadratic function given a vertex (, 4) and a y-intercept of Write an equation for b in terms of a. b Set the x-coordinate of the vertex equal to. a a Substitute for x a ( a ) Multiply both sides by a 4a b 4a Multiply both sides by 1 MUnit3Lesson /19/018

12 Example 5 (continued). Substitute the expression for b from Step 1, the coordinates of the vertex, and the y-intercept into the standard form of a quadratic equation. y a b c Standard form of a quadratic equation y a ( 4a) c Substitute 4a for b ( 4 ) a( ) ( 4a )( ) c Substitute the vertex (, 4) for x and y ( 4 ) a( ) ( 4a )( ) 4 Substitute the y-intercept of 4 for c 4 4a 8a 4 Simplify, then solve for a 8 4a a 3. Substitute the value of a into the equation for b from Step 1. b 4a Equation from Step 1 b 4( ) Substitute for a b 8 Simplify 4. Substitute a, b, and c into the standard form of a quadratic equation. y a b c Standard form of a quadratic equation y 8 4 Substitute for a, 8 for b, and 4 for c The equation of the quadratic function with a vertex of (, 4) and a y-intercept of 4 is y 8 4. MUnit3Lesson /19/018