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Quadratic Equations and Functions
Transcript of Quadratic Equations and Functions
5-1 Quadratic Function is a function that can be written in the standard form f(x)=ax +bx+c, where 'a' does NOT equal zero. 2 The graph of a quadratic function is called a parabola
Determine whether each function is linear or quadratic. Identify the quadratic, linear, and constant terms. f(x)=ax +bx+c 2 Quadratic
First Outer Inner Last
y = (2x + 3)(x – 4) First
y = (2x + 3)(x – 4) Outer
y = (2x + 3)(x – 4) Inner
y = (2x + 3)(x – 4) Last
y = 2x²-8x+3x-12
y = 2x²-5x-12 Things to Remember
The axis of symmetry is the line that divides a parabola into two parts that are mirror images.
The vertex of a parabola is the point at which the parabola intersects the axis of symmetry.
The y-value of the vertex of a parabola represents the maximum or minimum value of the function. Properties of Parabolas
5-2 The standard form of a quadratic function is y = ax2 + bx + c with a 0.When
b = 0, the function simplifies to y = ax2 + c.
The graph of y = ax2 + c is a parabola with an axis of symmetry x = 0, the y-axis.
The vertex of the graph is the y-intercept (0, c). Transforming Parabolas
5-3 To transform the graph of a quadratic function, you can use the
vertex form of a quadratic function, y = a(x - h)2 + k. Factoring Quadratic
5-4 Factoring is rewriting an expression as the product of its factors.
The greatest common factor (GCF) of an expression is a common factor of the
terms of the expression.It’s the common factor with the greatest coefficient and the
greatest exponent.You can factor any expression that has a GCF not equal to 1.
Finding Common Factors
Factor each expression.
For h . 0, the graph shifts right.
For h , 0, the graph shifts left.
For k . 0, the graph shifts up.
For k , 0, the graph shifts down.
The vertex is (h, k), and the axis of symmetry is the line x = h. Write y = 2x2 + 10x + 7 in vertex form.
x = Find the x-coordinate of the vertex.
= Substitute for a and b.
y = 2(-2.5)2 + 10(-2.5) + 7 Find the y-coordinate of the vertex.
The vertex is at (-2.5,-5.5).
y = a(x - h)2 + k Write the vertex form.
= 2(x - (-2.5))2 - 5.5 Substitute for a, h, and k.
= 2(x + 2.5)2 - 5.5 Simplify.
The vertex form of the function is y = 2(x + 2.5)2 - 5.5. Quadratic Equations
5-5 Zero Product Property: If ab=0, then a=0, or b=0
EX. If (x+3)(x-7)=0, then (x+3)=0 or (x-7)=0 Suppose you want to solve the equation
x2 + x – 20 = 0.
You can factor the left side as:
(x + 5)(x – 4) = 0
Now, by the zero product property, either
x + 5 = 0 or x – 4 = 0,
which means either x = –5 or x = 4. These are the two solutions of the equation. Complex Numbers
5-6 Complex Conjugate - The complex conjugate of a + bi is a - bi . When two complex conjugates are multiplied, the result is a2 + b2 .
Complex Number - A number of the form a + bi , where i =√-1 and a and b are real numbers.
Imaginary Number - A number of the form , where k < 0 , or a number of the form ki , where i = √-1
Imaginary Part - The term bi of a complex number a + bi Real Part - The term a of a complex number a + bi . Simplify square roots of negative numbers by factoring out √-1 = i and simplifying the resulting root.
EX- √-32 = √-1 . √32
=i . √2 . 2 . 2 . 2 . 2
= i . 4 . √2
=4 . i√2 √-100
= √-100 = √-1 . √100
= i . 10
10i Completing the Square
5-7 a perfect square trinomial is a polynomial that you get by squareing a binomial.(binomials are things like 'x + 3' or 'x − 5')
Example of perfect square trinomials (the red trinomials)
(x+ 1)² = x² + 2x + 1
Example of trinomials that are NOT perfect (red) square trinomials
(x+ 1)(x +2)= x² + 3x + 2 The Quadratic Formula
5-8 The solution of a quadratic equation is the value of x when you set the equation equal to zero
When you solve the following general equation: 0 = ax² + bx + c
Given a quadratic equation: ax ² + bx + c
The quadratic formula below will solve the equation for zero
The quadratic formula is : Graph y = x - 2x - 3. Label the vertex and the axis of symmetry.
Step 1 Find and graph the axis of symmetry.
Step 2 Find and graph the vertex.The x-coordinate of
the vertex is (1,0) , or 1.
The y-coordinate is y = (1) - 2(1) - 3 = -4.
So the vertex is (1,-4).
Step 3 Find and graph the y-intercept and its reflection.
Since c = -3, the y-intercept is (0,-3) and its
reflection is (2,-3).
Step 4 Evaluate the function for another value of x,
such as y = (3) - 2(3) - 3 = 0. Graph (3, 0)
and its reflection (-1, 0).
Step 5 Sketch the curve. 2 2 2