 # § 8.3 Quadratic Functions and Their Graphs. Blitzer, Intermediate Algebra, 4e – Slide #48 Graphing Quadratic Functions Graphs of Quadratic Functions The.

## Presentation on theme: "§ 8.3 Quadratic Functions and Their Graphs. Blitzer, Intermediate Algebra, 4e – Slide #48 Graphing Quadratic Functions Graphs of Quadratic Functions The."— Presentation transcript:

§ 8.3 Quadratic Functions and Their Graphs

Blitzer, Intermediate Algebra, 4e – Slide #48 Graphing Quadratic Functions Graphs of Quadratic Functions The graph of the quadratic function is called a parabola.

Blitzer, Intermediate Algebra, 4e – Slide #49 Graphing Quadratic Functions

Blitzer, Intermediate Algebra, 4e – Slide #50 Graphing Quadratic Functions Graphing Quadratic Functions With Equations in the Form T To graph 1) Determine whether the parabola opens upward or downward. If a > 0, it opens upward. It a < 0, it opens downward. 2) Determine the vertex of the parabola. The vertex is (h, k). 3) Find any x-intercepts by replacing f (x) with 0. Solve the resulting Quadratic equation for x. 4) Find the y-intercept by replacing x with 0. 5) Plot the intercepts and vertex and additional points as necessary. Connect these points with a smooth curve that is shaped like a cup.

Blitzer, Intermediate Algebra, 4e – Slide #51 Graphing Quadratic FunctionsEXAMPLE Graph the function SOLUTION We can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k. a = -2b = -4c = -8 1) Determine how the parabola opens. Note that a, the coefficient of, is -2. Thus, a < 0; this negative value tells us that the parabola opens downward.

Graphing Quadratic Functions 2) Find the vertex. The vertex of the parabola is at (h, k). Because h = -4 and k = -8, the parabola has its vertex at (-4, -8). CONTINUED 3) Find the x-intercepts. Replace f (x) with 0 in Find x-intercepts, setting f (x) equal to 0. Blitzer, Intermediate Algebra, 4e – Slide #52 Add to both sides. Divide both sides by 2. Apply the square root property. Simplify the radical. Subtract 4 from both sides.

Graphing Quadratic Functions Since no real solutions resulted from this step, there are no x- intercepts. CONTINUED 4) Find the y-intercept. Replace x with 0 in Blitzer, Intermediate Algebra, 4e – Slide #53 The y-intercept is -40. The parabola passes through (0,-40). 5) Graph the parabola. With a vertex at (-4,-8), no x-intercepts, and a y-intercept at -40, the graph of f is shown below. The axis of symmetry is the vertical line whose equation is x = -4.

Graphing Quadratic FunctionsCONTINUED Blitzer, Intermediate Algebra, 4e – Slide #54 y-intercept is: -40 Vertex: (-4,-8) Axis of symmetry: x = -4

Graphing Quadratic Functions Blitzer, Intermediate Algebra, 4e – Slide #55 The Vertex of a Parabola Whose Equation is T Consider the parabola defined by the quadratic function The parabola’s vertex is

Blitzer, Intermediate Algebra, 4e – Slide #56 Graphing Quadratic FunctionsEXAMPLE Graph the function Use the graph to identify its domain and its range. SOLUTION 1) Determine how the parabola opens. Note that a, the coefficient of, is 1. Thus, a > 0; this positive value tells us that the parabola opens upward. 2) Find the vertex. We know that the x-coordinate of the vertex is We identify a, b, and c for the given function. Note that a = 1, b = -4, and c = 6.

Blitzer, Intermediate Algebra, 4e – Slide #57 Graphing Quadratic Functions Substitute the values of a and b into the equation for the x- coordinate: CONTINUED The x-coordinate for the vertex is 2. We substitute 2 for x in the equation of the function to find the corresponding y-coordinate. The vertex is (2,2). 3) Find the x-intercepts. Replace f (x) with 0 in the original function. We obtain. This equation cannot be solved by factoring. We will use the quadratic formula instead.

Blitzer, Intermediate Algebra, 4e – Slide #58 Graphing Quadratic FunctionsCONTINUED Clearly, the discriminant is going to be negative, 16 – 24 = -8. Therefore, there will be no x-intercepts for the graph of the function. 4) Find the y-intercept. Replace x with 0 in the original function. The y-intercept is 6. The parabola passes through (0,6).

Blitzer, Intermediate Algebra, 4e – Slide #59 Graphing Quadratic FunctionsCONTINUED 5) Graph the parabola. With a vertex of (2,2), no x-intercepts, and a y-intercept at 6, the graph of f is shown below. The axis of symmetry is the vertical line whose equation is x = 2. Vertex: (2,2) y-intercept is: 6 Axis of symmetry: x = 2 Domain: All real numbers Range: All real numbers greater than or equal to 2.

Blitzer, Intermediate Algebra, 4e – Slide #60 Graphing Quadratic FunctionsCONTINUED Now we are ready to determine the domain and range of the original function. We can use the second parabola from the preceding page to do so. To find the domain, look for all inputs on the x-axis that correspond to points on the graph. To find the range, look for all the outputs on the y-axis that correspond to points on the graph. Looking at the first parabola from the preceding page, we see the parabola’s vertex is (2,2). This is the lowest point on the graph. Because the y-coordinate of the vertex is 2, outputs on the y-axis fall at or above 2.

Blitzer, Intermediate Algebra, 4e – Slide #61 Minimums & Maximums Minimum and Maximum: Quadratic Functions Consider 1) If a > 0, then f has a minimum that occurs at This minimum value is 2) If a < 0, then f has a maximum that occurs at This maximum value is

Blitzer, Intermediate Algebra, 4e – Slide #62 Minimums & MaximumsEXAMPLE A person standing close to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function models the ball’s height above the ground, s (t), in feet, t seconds after it was thrown. How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.

Blitzer, Intermediate Algebra, 4e – Slide #63 Minimums & MaximumsSOLUTION CONTINUED It might first be useful to have some sort of picture representing the situation. Below is some sort of picture. Point of interest

Blitzer, Intermediate Algebra, 4e – Slide #64 Minimums & MaximumsCONTINUED When the ball is released, it is at a height of 200 feet. That is, when the ball is released, the value of s (t) = 200. By the same token, when the ball finally hits the ground, it will of course be 0 feet above the ground. That is, when the ball hits the ground, the value of s (t) = 0. Therefore, to determine for what value of t the ball hits the ground, we replace s (t) with 0 in the original function. This is the given function. Replace s (t) with 0. Factor -8 out of all terms. Divide both sides by -8.

Blitzer, Intermediate Algebra, 4e – Slide #65 Minimums & MaximumsCONTINUED We will use the quadratic formula to solve this equation. Since time cannot be a negative quantity, the answer cannot be - 2.05 seconds. Therefore, the ball hits the ground after 6.05 seconds (to the nearest tenth of a second).

Blitzer, Intermediate Algebra, 4e – Slide #66 Minimums & Maximums Strategy for Solving Problems Involving Maximizing or Minimizing Quadratic Functions 1) Read the problem carefully and decide which quantity is to be maximized or minimized. 2) Use the conditions of the problem to express the quantity as a function in one variable. 3) Rewrite the function in the form 4) Calculate. If a > 0, f has a minimum at. This minimum value is. If a < 0, f has a maximum at. This maximum value is. 5) Answer the question posed in the problem.

Blitzer, Intermediate Algebra, 4e – Slide #67 Minimums & MaximumsEXAMPLE Among all pairs of numbers whose sum is 20, find a pair whose product is as large as possible. What is the maximum product? SOLUTION 1)Decide what must be maximized or minimized. We must maximize the product of two numbers. Calling the numbers x and y, and calling the product P, we must maximize P = xy. 2) Express this quantity as a function in one variable. In the formula P = xy, P is expressed in terms of two variables, x and y. However, because the sum of the numbers is 20, we can write x + y = 20.

Blitzer, Intermediate Algebra, 4e – Slide #68 Minimums & Maximums We can solve this equation for y in terms of x, substitute the result into P = xy, and obtain P as a function of one variable. CONTINUED y = 20 - xSubtract x from both sides of the equation: x + y = 20. Now we substitute 20 – x for y in P = xy. P = xy = x(20 – x). Because P is now a function of x, we can write P (x) = x(20 – x). 3) Write the function in the form. We apply the distributive property to obtain

Blitzer, Intermediate Algebra, 4e – Slide #69 Minimums & MaximumsCONTINUED 4) Calculate. If a < 0, the function has a maximum at this value. The voice balloons show that a = -1 and b = 20. P (x) = (20 – x)x = 20x - b = 20 a = -1 This means that the product, P, of two numbers who sum is 20 is a maximum when one of the numbers, x, is 10.

Blitzer, Intermediate Algebra, 4e – Slide #70 Minimums & MaximumsCONTINUED 5) Answer the question posed by the problem. The problem asks for the two numbers and the maximum product. We found that one of the numbers, x, is 10. Now we must find the second number, y. The number pair whose sum is 20 and whose product is as large as possible is 10, 10. The maximum product is 10 x 10 = 100. y = 20 – x = 20 – 10 = 10.

Blitzer, Intermediate Algebra, 4e – Slide #71 Minimums & MaximumsEXAMPLE You have 200 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed? SOLUTION 1) Decide what must be maximized or minimized. We must maximize area. What we do not know are the rectangle’s dimensions, x and y. xx y

Blitzer, Intermediate Algebra, 4e – Slide #72 Minimums & MaximumsCONTINUED 2) Express this quantity as a function in one variable. Because we must maximize area, we have A = xy. We need to transform this into a function in which A is represented by one variable. Because you have 200 feet of fencing, the sum of the lengths of the three sides of the rectangle that need to be fenced is 200 feet. This means that 2x + y = 200. We can solve this equation for y in terms of x, substitute the result into A = xy, and obtain A as a function in one variable. We begin by solving for y. y = 200 – 2x Subtract 2x from both sides.

Blitzer, Intermediate Algebra, 4e – Slide #73 Minimums & MaximumsCONTINUED Now we substitute 200 – 2x for y in A = xy. A = xy = x(200 – 2x) This function models the area, A (x), of any rectangle whose perimeter is 200 feet (and one side is not counted) in terms of one of its dimensions, x. The rectangle and its dimensions are illustrated in the picture at the beginning of this exercise. Because A is now a function of x, we can write A (x) = x(200 – 2x).

Blitzer, Intermediate Algebra, 4e – Slide #74 Minimums & MaximumsCONTINUED 3) Write the function in the form. We apply the distributive property to obtain a = -2 b = 200 4) Calculate. If a < 0, the function has a maximum at this value. The voice balloons show that a = -2 and b = 200.

Blitzer, Intermediate Algebra, 4e – Slide #75 Minimums & MaximumsCONTINUED 5) Answer the question posed in the problem. We found that x = 50. The picture at the beginning of this exercise shows that the rectangle’s other dimension is 200 – 2x = 200 – 2(50) = 200 – 100 = 100 feet. The dimensions of the rectangle that maximize the enclosed area are 50 feet by 100 feet. The rectangle that gives the maximum area has an area of (50 feet) x (100 feet) = 5,000 square feet. This means that the area, A (x), of a rectangle with a “3-sided” perimeter 200 feet is a maximum when the lengths of the two sides that are the same, x, are 50 feet.