1 Quadratic Functions

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1 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page 1 of 8 1 Quadratic Functions In this unit we will learn many of the algebraic techniques used to work with the quadratic function fx) = ax + bx + c. 1.1 Completing the Square The algebra process called completing the square is the way that we can isolate the variable x in a general quadratic. You will see this process again in precalculus and calculus, and it is definitely necessary to solve certain kinds of problems. I want you to see it now and be able to work with it for simple cases. Completing the Square: 1. Factor so that there is just a 1 in front of the x term.. Identify the coefficient of the x term. 3. Take half of this coefficient and square, then add and subtract so you don t change the equation.. Factor the perfect square you have created. 5. Simplify. ax + bx + c = a x + ba ) x + c Step 1. factor) = a x + ba ) x + c Step. coefficient of x is b a ) = a x + b ) b ) ) b a x + + c Step 3. blue terms are adding zero) = a x + b ) b ) ) b a x + + c the red terms are a perfect square) = a x + b ] ) ) b + c Step. factor perfect square) = a x + b = a x + b = a x + b ] a b ] b + c ] b c The process of completing the square shows that fx) = ax + bx + c = a x + b ] b c ) + c Step 5. distribute the a) simplify) common denominator)

2 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page of 8 1. Solving Quadratic Equations: The Quadratic Formula To solve simple quadratic equation of the form x = constant, we can use the square root property. Square root property: Solution to x = a is x = ± a. The square root property makes sense if you consider factoring x = a: x a = a a x a = 0 x a ) = 0 addition principle) properties of radicals) x a)x + a) = 0 factor as a difference of squares) x a = 0 or x + a = 0 x = a or x = a zero factor property) We can now derive the quadratic formula, which is used to solve more general quadratic equations of the form ax + bx + c = 0. a x + b a x + b ] b c a ax + bx + c = 0 ] b c = 0 from completing the square) b + c = + b c addition principle) x + b ] 1 a = b c 1 a ] = b c x + b x + b Quadratic formula: x + b = ± b c x + b = ± b c b = b ± b c x = b ± b c Solution to ax + bx + c = 0 is x = b ± b c. division principle) square root principle) properties of radicals addition principle) common denominator) There are three possibilities for the solution, based on the sign of the quantity b c.

3 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page 3 of 8 For the equation ax + bx + c = 0, the discriminant is defined as b c. The equation ax + bx + c = 0 has 1. Two real-valued solutions if b c > 0.. One real-valued solution if b c = Two complex-valued solutions if b c < 0. The two complex-valued solutions will be complex conjugates. example Use completing the square to solve x + x + 1 = 0. x + 6x + 1 = 0 x + 3x + 1 = 0 Step 1. factor, or divide to ensure x has coefficient of 1) x 3x + 1 = 0 Step. coefficient of x is 3) ) 3 ) 3 x + 3x = 0 Step 3. blue terms are adding zero) ) 3 ) 3 x + 3x = 0 Step. red terms are perfect square) x + 3 ) 9 + = 0 Step. factor perfect square) x + 3 ) = x + 3 ) = 7 addition principle) x + 3 = ± 7 square root property) x = 3 ± 7 As you can see, you can now solve quadratic equations that would not be ones that you could factor using our previous factoring techniques. You can use the quadratic formula to solve quadratic equations that previously you solved by factoring.

4 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page of 8 example Solve y = y yy + 1) + 3 y y = y + 1 yy + 1) = multiply by the LCD 15yy + 1)) y y y + 1) cancel common factors) yy + 1) + 5y + 1) = 60y multiply out) y + y + 5y + 5 = 60y y 1y + 5 = 0 y = b ± b c y = 1 ± 1) 1)5) 1) y = 1 ± 16 = 1 ± = 7 ± = 9 or y = 5 collect like terms) use quadratic formula) Since neither y = 9 nor y = 5 makes the LCD zero, these are both solutions. You could also substitute back into the original equation to determine that they are both solutions. 1.3 Pythagorean Theorem The hypotenuse in a right triangle is the side opposite the 90 degree angle in the triangle. Pythagorean theorem: a = b + c, where a is the length of the hypotenuse in a right triangle and b and c are the lengths of the other two sides. example A right triangle has one side of length cm and hypotenuse of length 5cm. What is the length of the other side? a = b + c 5) = ) + c 9 = c c = 3 choose positive root The other side has length 3 cm.

5 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page 5 of 8 1. Transforming into a Quadratic This is about using the mathematical concept of change of variables sometimes called substitution) which is a powerful concept and will be useful in the future. All solutions should begin by clearly identifying the change of variables that converts the equation into a quadratic equation. example Solve x + x) x + x) 1 = 0. Identify the substitution u = x + x. The equation then becomes x + x) x + x) 1 = 0 substiutution) u u 1 = 0 u + 3)u ) = 0 factor) u + 3 = 0 or u = 0 u = 3 or u = For each of these value of u we get a quadratic in x to solve: u = x + x = 3 x + x + 3 = 0 x = b ± b c x = ± ) 1)3) 1) x = ± 8 x = ± i 8 x = ± i x = 1 ± i zero factor property) quadratic formula) u = x + x = x + x = 0 x = b ± b c x = ± ) 1) ) 1) x = ± 0 x = ± 5 x = 1 ± 5 quadratic formula) There are four solutions to the original equation. Two are complex valued x = 1 ± i ), two are real valued x = 1 ± 5).

6 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page 6 of 8 Sketching Quadratics For a quadratic function y = fx) = ax + bx + c, we can create a sketch by determining four things: 1. x-intercepts: determine these by using the quadratic formula x intercept = b ± b c. NOTE: If b c < 0 then there are no x-intercepts they are not real numbers).. Vertex: x Vertex, y Vertex ) = b, f b )). x Vertex = b y Vertex = f b ) = b c How to remember this: The x-coordinate of the vertex will be right in the middle of the two x-intercepts, even if the x-intercepts are not real numbers! x Vertx = 1 b + b c + b ) b c b + b c b ) b c = 1 = 1 ) b = b 3. y-intercept: evaluate at x = 0, so figure out what f0) is.. The function opens up if a > 0, and opens down if a < 0. All this boils down nicely if you look at the completing the square version of the quadratic: fx) = ax + bx + c = a x + b ] b c standard form) = a x h] + k vertex form) from completing the square) The final form is called the vertex form since the vertex of the quadratic is h, k). Let s verify that. The smallest or largest value of the quadratic will be where x Vertex = b/) = h, since fh) = a 0 h h] + k = k = b c = y Vertex

7 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page 7 of 8 Sketching Quadratic in standard form fx) = ax + bx + c 1. Determine any x-intercepts if they exist) by solving fx) = 0,. Determine the vertex using x Vertex = b/) and y Vertex = fx Vertex ). 3. Determine the y-intercept, f0).. Determine if it opens up a > 0) or down a < 0), Sketching Quadratic in vertex form form fx) = ax h) + k 1. Determine any x-intercepts if they exist) by solving fx) = 0,. Determine the vertex using x Vertex = h and y Vertex = k. 3. Determine the y-intercept, f0).. Determine if it opens up a > 0) or down a < 0), example Sketch the parabola y = 5x + x 1. Label the vertex, y-intercept, and any x-intercepts on your sketch. To get the x-intercepts, use the quadratic formula. In this case, a = 5, b =, and c = 1. x = b ± b c = ± 5) 1) 5) = ± 56 ± 16 = + 16 = or = 1 or 0 = 6 5 or 16 To get the y-intercept, evaluate f0): y = f0) = 50) + 0) 1 = 1 A quadratic opens up if a > 0, and opens down if a < 0. Since a = 5 in this case, this quadratic opens up. You can now put this all together to get the sketch: The vertex is located at: x = b = 5) = 5 y = f ) 5 = 5 + 5) ) 1 5 = 6 5 The vertex is at ) 5, 6. 5

8 Unit 1 Quadratic Functions Lecture Notes Introductory Algebra Page 8 of 8 example What is the domain of the function fx) = x 3x + 1? Sketch your answer on a number line. Domain: what real numbers can I put into this expression and get a real number out? x 3x We only know how to algebraically solve linear inequalities, but we can sketch y = x 3x + 1 and get the answer from our sketch by figuring out where x 3x + 1 is positive. Since a = 1 > 0, this quadratic opens up. To get the x-intercepts, use the quadratic formula, a = 1, b = 3, and c = 1: x = b ± b c = 3 ± 3) 1)1) 1) = 3 ± 5 = or 3 5 Then the vertex is located at: x = b = 3) = 3 1) ) ) 3 3 y = = = = 5 ) 3 The vertex is at, 5. We don t need the y-intercept to answer the question. You can now put this all together to get the sketch:

Solving Quadratic Equations

Concepts: Solving Quadratic Equations, Completing the Square, The Quadratic Formula, Sketching Quadratics Solving Quadratic Equations Completing the Square ax + bx + c = a x + ba ) x + c Factor so the