# 6 1 Graphing Quadratic Functions Quadratic function equation

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6. 1: Graphing Quadratic Functions • Quadratic function: equation in the following form: ax 2 + bx + c = 0; a ≠ 0 • Graph is called a parabola

Graph by making a table of values.

Graph Answer: by making a table of values. x f(x) – 2 – 1 0 1 2 4 1 2 7 16

Parts of a quadratic function • Axis of symmetry: – A line where the parabola can be folded and be the exact same thing on both sides • Equation:

• Vertex – The point where the axis of symmetry meets the parabola • X – coordinate : • Y – intercept – Where the parabola crosses the y – axis • The value of c

Consider the quadratic function Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex.

Consider the quadratic function a. Find the y-intercept, the equation of the axis of symmetry, and the xcoordinate of the vertex. b. Make a table of values that includes the vertex. c. Use the information to graph the function.

Maximum/Minimum • The y – coordinate of a quadratic function is the maximum value or minimum value – If a > 0 (positive), then the parabola opens up and has a minimum value – If a < 0 (negative), then the parabola opens down and has a maximum value

Consider the function Determine whether the function has a maximum or a minimum value. State the maximum or minimum value of the function.

Consider the function a. Determine whether the function has a maximum or a minimum value. Answer: minimum b. State the maximum or minimum value of the function. Answer: – 5

6. 2: Solving Quadratic Equations by Graphing • The solutions of a quadratic equation are called roots of the equation • One method for finding the roots of a quadratic equation is to find the zeros of the function; meaning where y = 0

Solve by graphing.

Solve Answer: – 3 and 1 by graphing.

Number of Solutions • One Real Solution • Two Real Solutions • No Real Solutions

Solve by graphing.

Solve Answer: 3 by graphing.

Number Theory Find two real numbers whose sum is 4 and whose product is 5 or show that no such numbers exist.

Number Theory Find two real numbers whose sum is 7 and whose product is 14 or show that no such numbers exist. Answer: no such numbers exist

Estimate Solutions • Sometimes exact solutions (whole numbers) can not be found – Here we say what two numbers the solution is between

Solve by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located.

Solve by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. Answer: between 0 and 1 and between 3 and 4

Classwork

3/24/2009: Warm - Up

Homework Review Section 6. 1: 1. 15; x = 4; 4 2. 12; x = -2; -2 3. 1; x = 0. 5; 0. 5 4. Min; -9 5. Min; 5 6. Max; -8 7. Min; -8 8. Max; 3 9. Max; 0 Section 6. 2: 1. -1; 1 2. No real solutions 3. 1; 2 4. b/t 0 & 1; b/t -4 & -3 5. -6; -4 6. b/t -2 & -1; 2

Homework Review

6. 3: Solving Quadratic Equations by Factoring • Remember: • You have a quiz tomorrow on 6. 1 – 6. 3 • You will have an assignment at the end of the lesson that will be practice for this quiz • It will be taken up for a grade at the end of class

Zero Product Property • Another way to solve quadratic equations is by factoring. • Zero Product Property: For any numbers, if a ∙ b = 0, then one or both numbers have to be 0 • We will factor quadratic equations, then set all factors = 0 and solve them. (Always check!)

Solve by factoring.

Solve x 2 = 6 x by factoring.

Solve x 2 + 6 x – 16 = 0 by factoring.

Solve x 2 + 9 = 6 x by factoring.

Solve by factoring.

Solve 2 x 2 + 7 x = 15 by factoring.

Solve each equation by factoring. a. Answer: {0, 3} b. Answer: c. x 2 – 4 x = 21 Answer: {-3, 7}

Double Roots • When the quadratic equation has only one solution (graph only touches the x – axis once or kisses it) then we have what is called a double root (because it occurs twice) • When we factor and our factors are the same, this means there is a double root, and only one real solution • We can always check ourselves by looking at the graph of the quadratic equation

Solve by factoring.

Solve Answer: {– 5} by factoring.

Multiple-Choice Test Item What is the positive solution of the equation ? A – 3 B 5 C 6 D 7

Multiple-Choice Test Item What is the positive solution of the equation ? B – 5 A 5 C 2 D 6 Answer: C

Writing quadratic equations • You may be given solutions and asked to write the quadratic equation • You do this by working backwards: • • Set x = to the solution(s) Move the solution(s) to where the equation = 0 Multiply (FOIL or BOX) the equation(s) Simplify as needed

Write a quadratic equation with equation in quadratic form. and 6 as its roots. Write the

Write a quadratic equation with equation in quadratic form. Answer: and 5 as its roots. Write the

Classwork • The following assignment needs to be completed by the end of class as a review for your quiz tomorrow. – Once you have finished this you are done for the day • Homework: – Study 6. 1 – 6. 3 – Quiz first thing tomorrow

3/25/2009: Warm - Up

Quiz Review

Quiz Time • You will have 30 – 35 minutes to complete this quiz • Take your time • Good Luck! • We will begin 6. 5 once everyone has finished – we will do 6. 4 tomorrow

6. 5: The Quadratic Formula and the Discriminant • Just like graphing and factoring, the quadratic formula can give us the solution to a quadratic equation (proof pg. 313) • Quadratic Formula: • Just like before you can check yourself by looking at the graph

Solve by using the Quadratic Formula.

Solve x 2 – 12 x = 28 by using the Quadratic Formula.

Solve Answer: 2, – 15 by using the Quadratic Formula.

Solve by using the Quadratic Formula. Identify a, b, and c. Then, substitute these values into the Quadratic Formula.

Solve Formula. Answer: 11 by using the Quadratic

Solve by using the Quadratic Formula.

Solve Answer: by using the Quadratic Formula. or approximately 0. 7 and 4. 3

Solve by using the Quadratic Formula. Write the equation in the form

Solve Answer: by using the Quadratic Formula.

Roots and the Discriminant • Discriminant: b 2 – 4 ac • What is under the radical • The discriminant determines: • How many roots you have • One or Two • What type of roots you have • Real, Rational, Irrational, Complex

Value of Discriminant Type/Number of Roots Discriminant is positive and a perfect square 2 real, rational roots Discriminant is positive and not a perfect square 2 real, irrational roots Discriminant is zero 1 real, rational root Discriminant is negative 2 complex roots Example of a Graph

Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Find the value of the discriminant for. Then describe the number and type of roots for the equation.

Find the value of the discriminant for Then describe the number and type of roots for the equation. .

Find the value of the discriminant for each quadratic equation. Then describe the number and type of roots for the equation. a. Answer: 0; 1 rational root b. Answer: – 24; 2 complex roots c. Answer: 5; 2 irrational roots d. Answer: 64; 2 rational roots

Which method do I use? • If the problem does not tell me which method to use, how do I know which on is the easiest, quickest, and most efficient method to use?

Solving Quadratic Equations (pg 317) Method Can be Used When to Use Graphing Sometimes If an exact answer is not required; best to use to check yourself Factoring Sometimes If c = 0 or factors are easy to see Square Root Property Sometimes If an equation that is a perfect square is equal to a constant Completing the Square Always If b is an even number Quadratic Formula Always If all other methods fail or are too tedious

Classwork/Homework

Solve Root Property. by using the Square Original equation Factor the perfect square trinomial. Square Root Property Subtract 7 from each side. or Write as two equations. Solve each equation. Answer: The solution set is {– 15, 1}.

Solve Root Property. Answer: {3, 13} by using the Square

Solve Root Property. by using the Square Original equation Factor the perfect square trinomial. Square Root Property Add 5 to each side. or Write as two equations. Use a calculator.

Answer: The exact solutions of this equation are and The approximate solutions are 1. 5 and 8. 5. Check these results by finding and graphing the related quadratic function. Original equation Subtract 12 from each side. Related quadratic function

Check Use the ZERO function of a graphing calculator. The approximate zeros of the related function are 1. 5 and 8. 5.

Solve Root Property. Answer: by using the Square

Find the value of c that makes a perfect square. Then write the trinomial as a perfect square. Step 1 Find one half of 16. Step 2 Square the result of Step 1. Step 3 Add the result of Step 2 to Answer: The trinomial as can be written

Find the value of c that makes a perfect square. Then write the trinomial as a perfect square. Answer: 9; (x + 3)2

Solve by completing the square. Notice that is not a perfect square. Rewrite so the left side is of the form Since add 4 to each side. Write the left side as a perfect square by factoring.

Square Root Property Subtract 2 from each side. or Write as two equations. Solve each equation. Answer: The solution set is {– 6, 2}.

Solve Answer: {– 6, 1} by completing the square.

Solve by completing the square. Notice that is not a perfect square. Divide by the coefficient of the quadratic term, 3. Add to each side. Since add to each side.

Write the left side as a perfect square by factoring. Simplify the right side. Square Root Property Add to each side.

or Write as two equations. Solve each equation. Answer: The solution set is

Solve Answer: by completing the square.

Solve by completing the square. Notice that is not a perfect square. Rewrite so the left side is of the form Since add 1 to each side. Write the left side as a perfect square by factoring.

Square Root Property Subtract 1 from each side. Answer: The solution set is Notice that these are imaginary solutions.

Check A graph of the related function shows that the equation has no real solutions since the graph has no x-intercepts. Imaginary solutions must be checked algebraically by substituting them in the original equation.

Solve Answer: by completing the square.

Example 1 Graph a Quadratic Function in Vertex Form Example 2 Write y = x 2 + bx + c in Vertex Form Example 3 Write y = ax 2 + bx + c in Vertex Form, a 1 Example 4 Write an Equation Given Points

Analyze Then draw its graph. h = 3 and k = 2 Answer: The vertex is at (h, k) or (3, 2) and the axis of symmetry is The graph has the same shape as the graph of but is translated 3 units right and 2 units up. Now use this information to draw the graph.

Step 1 Plot the vertex, (3, 2). Step 2 Draw the axis of symmetry, Step 3 Find and plot two points on one side of the axis of symmetry, such as (2, 3) and (1, 6). Step 4 Use symmetry to complete the graph. (1, 6) (5, 6) (2, 3) (3, 2) (4, 3)

Analyze Then draw its graph. Answer: The vertex is at (– 2, – 4), and the axis of symmetry is The graph has the same shape as the graph of ; it is translated 2 units left and 4 units down.

Write the function. in vertex form. Then analyze Notice that is not a perfect square. Complete the square by adding Balance this addition by subtracting 1.

Write as a perfect square. This function can be rewritten as So, and Answer: The vertex is at (– 1, 3), and the axis of symmetry is Since the graph opens up and has the same shape as but is translated 1 unit left and 3 units up.

Write the function. in vertex form. Then analyze Answer: vertex: (– 3, – 4); axis of symmetry: opens up; the graph has the same shape as the graph of but it is translated 3 units left and 4 units down.

Write in vertex form. Then analyze and graph the function. Original equation Group and factor, dividing by a. Complete the square by adding 1 inside the parentheses. Balance this addition by subtracting – 2(1).

Write as a perfect square. Answer: The vertex form is So, and The vertex is at (– 1, 4) and the axis of symmetry is Since the graph opens down and is narrower than It is also translated 1 unit left and 4 units up. Now graph the function. Two points on the graph to the right of are (0, 2) and (0. 5, – 0. 5). Use symmetry to complete the graph.

Write in vertex form. Then analyze and graph the function. Answer: vertex: (– 1, 7); axis of symmetry: x = – 1; opens down; the graph is narrower than the graph of y = x 2, and it is translated 1 unit left and 7 units up.

Write an equation for the parabola whose vertex is at (1, 2) and passes through (3, 4). The vertex of the parabola is at (1, 2) so and Since (3, 4) is a point on the graph of the parabola , and Substitute these values into the vertex form of the equation and solve for a. Vertex form Substitute 1 for h, 2 for k, 3 for x, and 4 for y. Simplify.

Subtract 2 from each side. Divide each side by 4. Answer: The equation of the parabola in vertex form is

Check A graph of verifies that the parabola passes through the point at (3, 4).

Write an equation for the parabola whose vertex is at (2, 3) and passes through (– 2, 1). Answer:

Example 1 Graph a Quadratic Inequality Example 2 Solve ax 2 + bx + c 0 Example 3 Solve ax 2 + bx + c 0 Example 4 Write an Inequality Example 5 Solve a Quadratic Inequality

Graph Step 1 Graph the related quadratic equation, Since the inequality symbol is >, the parabola should be dashed.

Graph Step 2 Test a point inside the parabola, such as (1, 2). So, (1, 2) is a solution of the inequality. (1, 2)

Graph Step 3 Shade the region inside the parabola. (1, 2)

Graph Answer:

Solve by graphing. The solution consists of the x values for which the graph of the related quadratic function lies above the x-axis. Begin by finding the roots of the related equation. Related equation Factor. or Zero Product Property Solve each equation.

Sketch the graph of the parabola that has x-intercepts at 3 and 1. The graph lies above the x-axis to the left of and to the right of Answer: The solution set is

Solve Answer: by graphing.

Solve by graphing. This inequality can be rewritten as The solution consists of the x-values for which the graph of the related quadratic equation lies on and above the x-axis. Begin by finding roots of the related equation. Related equation Use the Quadratic Formula. Replace a with – 2, b with – 6 and c with 1.

or Simplify and write as two equations. Simplify. Sketch the graph of the parabola that has x-intercepts of – 3. 16 and 0. 16. The graph should open down since a < 0.

Answer: The graph lies on and above the x-axis at and between these two values. The solution set of the inequality is approximately

Check Test one value of x less than – 3. 16, one between – 3. 16 and 0. 16, and one greater than 0. 16 in the original inequality.

Solve by Answer: by graphing.

Sports The height of a ball above the ground after it is thrown upwards at 40 feet per second can be modeled by the function where the height h(x) is given in feet and the time x is in seconds. At what time in its flight is the ball within 15 feet of the ground? The function h(x) describes the height of the ball. Therefore, you want to find values of x for which Original inequality Subtract 15 from each side.

Graph the related function graphing calculator. using a The zeros are about 0. 46 and 2. 04. The graph lies below the x-axis when or

Answer: Thus, the ball is within 15 feet of the ground for the first 0. 46 second of its flight and again after 2. 04 seconds until the ball hits the ground at 2. 5 seconds.

Sports The height of a ball above the ground after it is thrown upwards at 28 feet per second can be modeled by the function where the height h(x) is given in feet and the time x is given in seconds. At what time in its flight is the ball within 10 feet of the ground? Answer: The ball is within 10 feet of the ground for the first 0. 5 second of its flight and again after 1. 25 seconds until the ball hits the ground.

Solve algebraically. First, solve the related equation . Related quadratic equation Subtract 2 from each side. Factor. or Zero Product Property Solve each equation.

Plot – 2 and 1 on a number line. Use closed circles since these solutions are included. Notice that the number line is separated into 3 intervals.

Test a value in each interval to see if it satisfies the original inequality.

Answer: The solution set is This is shown on the number line below.

Solve Answer: algebraically.

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