Graphing Quadratic Functions Rational Functions Conic Sections Chapter 11 Graphing Quadratic Functions Rational Functions Conic Sections 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8

Quadratic Functions and Their Graphs 11.1 Quadratic Functions and Their Graphs Objective: Graph quadratic functions of the form f(x) = a(x – h)2 + k

We first examined the graph of y = x2 back in chapter 3 We first examined the graph of y = x2 back in chapter 3. Then wee looked at the graphs of general quadratic functions of the form f(x) = ax2 + bx + c. We discovered that the graph of a quadratic function is a parabola opening upward or downward, depending on the coefficient of the x2 term, a. The highest point or lowest point on the parabola is the vertex. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola.

Graphing the Parabola Defined by f(x) = x2 + k If k is positive, the graph of y = x2 is shifted If k is negative, the graph of y = x2 is shifted The vertex is (0, k), and the axis of symmetry is the y-axis. up k units. down │k│ units. Graphing the Parabola Defined by f(x) = (x – h)2 If f(x) = (x – h)2, the graph of y = x2 is shifted If f(x) = (x + h)2 , the graph of y = x2 is shifted The vertex is (h, 0), and the axis of symmetry is the x = h. to the right h units. to the left |h| units.

Graphing the Parabola Defined by f(x) = (x – h)2 + k The parabola has the same shape as y = x2. The vertex is (h, k), and the axis of symmetry is the vertical line x = h.

Example 1 f(x) = x2 Vertex: Axis of sym: g(x) = x2 + 3 Vertex: (0,0) x y x = 0 g(x) = x2 + 3 Vertex: Axis of sym: (0,3) x = 0 h(x) = x2 – 3 Vertex: Axis of sym: (0,–3) x = 0

Example 2 Given the graph f(x) = x2. Graph g(x) = (x – 3)2 and h(x) = (x + 3)2. f(x) = x2 g(x) = (x – 3)2 Vertex: Axis: Vertex: (3, 0) Axis: x = 3 h(x) = (x + 3)2 Vertex: Axis: Vertex: (3, 0) Axis: x = 3

Example 3 Graph g(x) = (x – 2)2 + 4 f(x) = x2 g(x) = (x – 2)2 + 4 y g(x) = (x – 2)2 + 4 Vertex: Axis: (2, 4) x = 2

Graph g(x) = 3x2 and h(x) = (1/3)x2 Example 4 Graph f(x) = x2 Graph g(x) = 3x2 and h(x) = (1/3)x2 How do the shapes of the graphs compare? x y f(x) = x2 g(x) = 3x2 h(x) = (1/3)x2 Time out on the whiteboard: graphing with a steepness of ½, ¼, 2/5, …

Graphing the Parabola Defined by f(x) = ax2 If a is positive, the parabola open upward, and if a is negative, the parabola opens downward. If |a| > 1, the graph of the parabola is narrower than the graph of y = x2. If |a| < 1, the graph of the parabola is wider than the graph of y = x2.

Example 5 Graph g(x) = –4(x + 2)2 – 1. Find the vertex and axis of symmetry. Vertex: Axis of sym: The graph opens: The steepness is: The graph is shifted: (–2, –1) x = –2 down (reflected over x-axis) 4 times as steep as y= x2. 2 units left and one unit down

11.1 summary

Further Graphing of Quadratic Functions 11.2 Further Graphing of Quadratic Functions Objectives: Write quadratic functions in the form f(x) = a(x – h)2 + k Derive a formulas for finding the vertex of a parabola Find the minimum or maximum values of a quadratic function

If the quadratic is written in the form f(x) = a(x – h)2 + k, then we can find the vertex, axis of symmetry, whether it opens up or down, and the width.

Find the vertex, axis, and any intercepts. Graph f(x) = 3x2 + 6x + 4. Find the vertex, axis, and any intercepts. Example 1 x y From f(x) = -3x2+ 6x + 4 Axis of sym: x = –b/2a x = 1 Vertex: f(1) = -3(1)2+ 6(1) + 4 = 7 (1,7) Re-write the equation into f(x) = a(x – h)2 + k form f(x) = –3(x – 1)2 + 7 = –6/-6 Calculate X- & Y-intercept on next slide:

y-intercept is the point (0, 4). Graph f(x) = 3x2 + 6x + 4. Find the vertex, axis, and any intercepts. Vertex: Axis of sym: (1, 7) x-intercept(s): find x when y = 0 0 = –3(x – 1)2 + 7 3(x – 1)2 = 7 (x – 1)2 = 7/3 x = 1 y-intercept: find y when x = 0. y = –3(0)2 + 6(0) + 4 y = 0 + 0 + 4 y = 4 y-intercept is the point (0, 4). x-intercepts are ≈ (−0.7, 0) and (2.7, 0) continue

Example 1 Graph f(x) = 3x2 + 6x + 4. Find the vertex, axis of symmetry, and any intercepts. Complete the square on x to write the equation in the form y = a(x – h)2 + k.

Vertex Formula The graph of f(x) = ax2 + bx + c, when a ≠ 0, is a parabola with vertex

Example 2 The Utah Ski Club sells calendars to raise money. The profit P, in cents, from selling x calendars is given by the equation P(x) = 360x – x2. What is the maximum profit the club will earn? a = 1 and b = 360 Since the maximum value will occur at the vertex, we find the coordinates according to the previous formula. The maximum profit will occur when the club sells 180 calendars. We substitute that number into the profit formula to find P(180) = 360(180) – (180)2 = 64800 – 32400 = 32400 cents = $324

Example 3: Given y = ¼ (x – 3)2 + 2 Find the vertex, axis of symmetry, and make a table of points used to graph the parabola. Vertex: (3, 2) Axis of Symmetry: x = 3 Table of points: X Y 3 2 –1 6 7 6

Example 4: Write the equation of the parabola in vertex form given it has a vertex of (4, –4) and x-intercepts of 2 and 6. Vertex Form of a Parabola: y = a(x – h)2 + k The vertex (4, –4): y = a(x – 4)2 – 4 X intercepts of 2 and 6: means the parabola passes through the points (2,0) and (6,0) Choose one of the xint: 0 = a(2 – 4)2 – 4 I choose (2,0) 4 = a(4) 1 = a Answer: y = (x – 4)2 – 4

11.2 Summary Objectives: Write quadratic functions in the form f(x) = a(x – h)2 + k Derive a formulas for finding the vertex of a parabola Find the minimum or maximum values of a quadratic function (this is the y-coordinate of the vertex)

Graphing Rational Functions by Transformations 11.3 Graphing Rational Functions by Transformations Objectives: Find the domain of rational functions Graph f(x) = 1/x and f(x) = 1/x2 Identify vertical and horizontal asymptotes. Use transformations to graph rational functions.

Rational functions are quotients of polynomial functions Rational functions are quotients of polynomial functions. This means that rational functions can be expresses as where p and q are polynomial functions and q(x) ≠ 0. The domain of a rational function consists of all real numbers except the x-values that make the denominator zero.

Example 1 Find the domain of the rational functions: Domain: we can NEVER divide by 0 so x – 4 ≠ 0 x ≠ 4 Domain: all real #’s except 4 Domain: we can NEVER divide by 0 so x2 – 36 ≠ 0 x ≠ ±6 Domain: all real #’s except 6 or –6

The most basic rational function is the reciprocal function, defined by f(x) = 1/x2. The denominator of the reciprocal function is zero when x = 0, so the domain of f is the set of all real numbers except 0. 1 1 ¼ ¼ 1/9 1/9 4 4 Vertical asymptote: x = 0 Horizontal asymptote: y = 0

Locating Vertical Asymptotes Let be a rational function in which p(x) and q(x) have no common factors. If a is a zero of q(x), the denominator, then x = a is a vertical asymptote of the graph of f.

Example 2 Find the vertical asymptotes, if any, of the graphs of the rational function. Domain: x – 4 ≠ 0 x ≠ 4 Vertical Asy: x = 4 Domain: x + 6 ≠ 0, x – 6 ≠ 0 x ≠ –6, x ≠ 6 Vertical Asy: x = –6 and x = 6 Domain: x2 + 36 ≠ 0 x all Real #’s Vertical Asy: none

BOBO BETC BOTD Degree bigger on bottom → y = 0 Degree bottom equals top → y = coef BOTD Degree bigger on top → divide to get oblique asymptote (no horizontal asy)

Example 3 Find the horizontal asymptotes, if any, of the graph of the rational function. BOBO BETC BOTD BOBO BETC BOTD Horizontal Asy: y = 0 Horizontal Asy: y = 6

Example 4 Use the graph of

11.3 summary Objectives: Find the domain of rational functions denominator ≠ 0 Graph f(x) = 1/x and f(x) = 1/x2 you need to memorize these two parent functions Identify vertical solve denominator = 0 Identify horizontal asymptotes. BOBO BETC BOTD Use transformations to graph rational functions. horizontal shift: inside with the x backwards right/left vertical shift: outside at the end up/down

Further Graphing of Rational Functions 11.4 Further Graphing of Rational Functions Objectives: Graph rational functions Identify slant asymptotes Solve applied problems involving rational functions

Graph: Example 1 y-intercept. find y when x = 0 x y x-intercept(s). Find x when y = 0 2x – 3 = 0 x = 3/2 x-intercept = (3/2, 0) x y 2 3 4 5 6 -1 Vertical asy: x = 3 Horizontal asy: y = 2 asy 5 3.5 3

Graph: y-intercept = (0, 0) Example 2 y-intercept. find y when x = 0 y x-intercept(s). Find x when y = 0 2x2 = 0 x-intercept = (0, 0) x y -4 -3 -2 -1 1 2 3 4 4.57 Vertical asy: x = 3 and x = –3 Horizontal asy: y = 2 asy -1.6 -.25 -.25 -1.6 asy 4.57

The graph of a rational function has a slant asymptote if the degree of the numerator is one more than the degree of the denominator. The equation of the slant asymptote can be found by division.

Example 3 Find the slant asymptote. The degree of the numerator is one more than the degree of the denominator, and x – 3 is not a factor of the numerator, the graph has a slant asymptote. To find the equation of the slant asymptote divide x – 3 into x2 – 4x – 5. The slant asymptote is y = x – 1 3 1 –4 –5 –3 –1 -8

Graph: Example 4 y-intercept. find y when x = 0 y-intercept = (0, 5/3) x-intercept(s). Find x when y = 0 (x – 5)(x + 1) = 0 x = 5 or -1 x-intercept = (5,0), (-1,0) Vertical asy: x = 3 Horizontal asy: none Oblique asy: y = x – 1 x y 1 2 3 4 5 4 9 asy -5

11.4 Summary Objectives: Graph rational functions y-intercept: find y when x = 0 x-intercept: find x when y = 0 vertical asymptote: denominator = 0 a graph will have either a horizontal OR an oblique asymptote BOBO BETC BOTD

The Parabola and the Circle 11.5 The Parabola and the Circle Objectives: Graph parabolas of the form: x = a(y – k)2 + h and y = a(x – h)2 + k Graph circles of the form: (x – h)2 + (y – k)2 = r2 Write the equation of a circle, given its center and radius. Find the center and radius of a circle, given its equation.

Conic Sections Conic sections derive their name because each conic section is the intersection of a right circular cone and a plane. The circle is formed by the intersection of a plane perpendicular to the symmetry axis of the cone The parabola is formed by the intersection of a cone and a plane parallel with the generating line of the cone The ellipse is formed by the intersection of a cone and a plane that does not pass through the apex. The hyperbola is formed by the intersection of a cone and a plane parallel to the axis of the cone.

y = a(x – h)2 + k x = a(y – k)2 + h Parabolas Just as y = a(x – h)2 + k is the equation of a parabola that opens upward or downward, x = a(y – k)2 + h is the equation of a parabola that opens to the right or to the left. y = a(x – h)2 + k x = a(y – k)2 + h x y x y x y x y a > 0 (h, k) (h, k) (h, k) y = k a < 0 y = k (h, k) a < 0 a > 0 x = h x = h

Example 1 y = a(x – h)2 + k x = a(y – k)2 + h Graph the parabola x = 4y2 Opens to the right. a = 4, h = 0, k = 0 Vertex: (0, 0) Axis of symmetry: y = 0 x y

Example 2 y = a(x – h)2 + k x = a(y – k)2 + h Graph the parabola x = 3(y – 4)2 + 1. Opens to the right. Vertex: (1, 4) Axis of symmetry: y = 4 x y y = 4

Example 3 Graph the parabola y = x2 + 12x + 25. y = a(x – h)2 + k x = a(y – k)2 + h Axis of symmetry: x = -b/2a Axis of symmetry: x = -12/2 x = -6 Y-coordinate of vertex: y = (-6)2 + 12(-6) + 25 y = -11 Vertex: (-6,-11) Standard form: y = (x + 6)2 - 11

Example 4 y = a(x – h)2 + k x = a(y – k)2 + h Graph the parabola x = y2 – 8y + 17 Opens to the right. Axis of symmetry: y = 8/2 y = 4 X coordinate of vertex: x = (4)2 – 8(4) + 17 x = 1 Vertex (1,4) Standard form: x = (y – 4)2 + 1 x y y = 4 continue

A circle is the set of all points in a plane that are the same distance from a fixed point called the center. The distance is called the radius. Circle The graph of (x – h)2 + (y – k)2 = r2 is a circle with center (h, k) and radius r. y x (h, k) r

Graph x2 + y2 = 9. (x – 0)2 + (y – 0)2 = 32 The center The radius Example 5 Graph x2 + y2 = 9. (x – 0)2 + (y – 0)2 = 32 The center The radius Graph (x – 3)2 + y2 = 9. (x – 3)2 + (y – 0)2 = 32 The center The radius (0,0) (3,0) r = 3 r = 3 x y r = 3 (3, 0)

Example 6 Find the equation of the circle with center (–7, 6) and radius 2. h = –7, k = 6, and r = 2 (x – h)2 + (y – k)2 = r2 The equation can be written (x + 7)2 + (y – 6)2 = 4

Example 7 Graph x2 + y2 + 4x – 8y – 16 = 0 x2 + y2 + 4x – 8y – 16 = 0 To write the equation in standard form, group terms and complete the square on each variable. x2 + y2 + 4x – 8y – 16 = 0 (x2 + 4x + _) + (y2 – 8y + _) = 16 + _ + _ (x2 + 4x + 4) + (y2 – 8y + 16) = 16 + 4 + 16 (x + 2)2 + (y – 4)2 = 36 The center is (–2, 4) radius is 6.

Summary 11.5 UIL Math TEAM Parabola Circle y = a(x – h)2 + k Vertex: (h,k) x = a(y – k)2 + h Center: (h,k) Radius = r (x – h)2 + (y – k)2 = r2 UIL Math TEAM More on the next slides for you

The Ellipse and the Hyperbola 11.6 The Ellipse and the Hyperbola Circle Ellipse Parabola Hyperbola Conic sections derive their name because each conic section is the intersection of a right circular cone and a plane.

Definitions A Parabola is the set of all points in a plane that are the same distance from a given point, called the focus, and a given line, called the directrix. A Circle is the set of all points in a plane that are equidistant from a given point in the plane, called the center. An ellipse is the set of all points in a plane such that the sum of the distances from two given points in the plane, called foci, is constant. A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two given points in the plane, called foci, is constant.

The graph of an equation of the form Ellipse with Center (0, 0) The graph of an equation of the form is an ellipse with center (0, 0). The x-intercepts are (a, 0) and (–a, 0), and the y-intercepts are (0, b) and (0, –b). y x a b –b –a

Example 1 Graph center (0, 0) a = 2 and b = 3 x-intercepts: (2, 0) and (2, 0) y-intercepts: (0, 3) and (0, 3)

Example 2 Graph center (0, 0) a = 3 and b = 2 x-intercepts: (3, 0) and (3, 0) y-intercepts: (0, 2) and (0, 2)

Example 3 Graph Center: (3, 2) a = 5 and b = 6 Since a = 5 count 5 units right and 5 units left for two points. Since b = 6 count 6 units up and 6 units down for two points. x y

Hyperbola with Center (0, 0) x y a a The graph of an equation of the form is a hyperbola with center (0, 0) and x-intercepts (a, 0) and (–a, 0). x y b b The graph of an equation of the form is a hyperbola with center (0, 0) and y-intercepts (0, b) and (0, –b). x y The asymptotes of the hyperbola are dashed lines used to sketch the graph of the hyperbola.

Example Graph The equation is of the form so its graph is a hyperbola that opens to the left and right. x y It has center (0, 0) Vertices: right/left 3 units (3,0) and (-3,0) Box: right/left 3 units and up/down 6 units

Example Graph This hyperbola opens up/down Center: (4, -1) Vertices: up/down 3 units (4,2) and (4,-4) Box: up/down 3 units & right/left 4 units x y Equation of asymptotes: y + 1 = ± 3/4(x – 4)

What type of conic does each equation represent? x2 – y2 + 2x – 4y – 20 = 0 x2 + y2 + 2x – 4y – 20 = 0 x2 + 2y2 + 2x – 4y – 20 = 0 x2 + y + 2x – 20 = 0 x = y2 + 2x – 4y – 20 3x2 = –3y2 + 2x – 4y – 20 3x2 = 2y2 + 6x – 12y – 20 x2 = –4y2 + 2x – 4y – 20 Hyperbola Circle Ellipse Parabola

11.6 Ellipse Summary Center (h,k) and a > b The endpoints on the longer axis (major axes) are vertices The endpoints on the shorter axis (minor axis) are co-vertices

11.6 Hyperbola Summary center (h,k) b b a a Asy: y – k = ±a/b(x – h) Asy: y – k = ±b/a(x – h)

Definitions A Parabola is the set of all points in a plane that are the same distance from a given point, called the focus, and a given line, called the directrix. A Circle is the set of all points in a plane that are equidistant from a given point in the plane, called the center. An ellipse is the set of all points in a plane such that the sum of the distances from two given points in the plane, called foci, is constant. A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two given points in the plane, called foci, is constant. UIL: Mathematics add on Foci points and the value of c

Solving Nonlinear Systems of Equations 11.7 Solving Nonlinear Systems of Equations Objective: Solve a nonlinear system by substitution and elimination

A nonlinear system of equations is a system of equations at least one of which is not linear. The substitution method or the elimination method may be used to solve the system.

Example 1 Solve the system y = x – 2 y + 2 = -3 y + 2 = 2 x2 + y = 4 Let x = -3 Let x = 2 y + 2 = -3 y + 2 = 2 x2 + y = 4 y = -5 y = 0 The solutions are (–3, –5) &(2, 0) x2 + (x – 2) = 4 x y x2 + x – 6 = 0 Graph Y = X – 2 Y = -X2 + 4 (x + 3)(x – 2) = 0 x = -3 or x = 2

Solve the system Check both solutions in both equations. Check the solution (2, 0) Check the solution (–3,–5) x2 + y = 4 x2 + y = 4 22 + 0 = 4 (–3)2+ (–5) = 4 4 = 4 9 – 5 = 4 y + 2 = x y + 2 = x –5 + 2 = –3 0 + 2 = 2 –3 = –3 2 = 2

Example x2 + 2y2 = 4 Solve the system. –x2 + y2 = –4 3y2 = 0 y = 0 Mult (-1) 3y2 = 0 Check by graphing: y = 0 Sub 0 for y to find x. x2 + 2y2 = 4 x2 + 2(0)2 = 4 x2 = 4 x = 2 or x = –2 The solutions are (–2, 0) and (2, 0).

Add on: Combine like terms

Add on: Solve the system of equations: x2 + y2 = 1681 40x + 9y = 0   When x = 9, y = -40/9 (9) = -40 When x = -9, y = -40/9 (-9) = 40     Answer: (9, –40) and (–9, 40)  

11.7 summary Objective: Solve a nonlinear system by substitution and elimination

Nonlinear Inequalities and Systems of Inequalities 11.8 Nonlinear Inequalities and Systems of Inequalities Objectives: Graph a nonlinear inequality Graph a system of nonlinear inequalities

Example 1 Select a test point to determine which region contains the solutions. x y 4 – 4 – 3 3 (0, 0) True

Example 2 Region A x y The hyperbola divides the plane into three regions. Select a test point in each region to determine the solutions. (0, 4) Region B (0, 0) Region A Region B Region C (0, – 4 ) Region C False True False

Example 3 Graph the system. Graph each inequality on the same set of axes. The final answer is the overlap. Where is it true for both equations? x y

11.8 summary Objectives: Graph a nonlinear inequality Graph a system of nonlinear inequalities where does blue & yellow = green