 # THE GRAPH OF A QUADRATIC FUNCTION

## Presentation on theme: "THE GRAPH OF A QUADRATIC FUNCTION"— Presentation transcript:

THE GRAPH OF A QUADRATIC FUNCTION

f(x) = ax 2 + bx + c where a, b & c are real numbers and a  0 The domain of a quadratic function is all real numbers.

As we’ve already seen, f(x) = x2 graphs into a PARABOLA. This is the simplest quadratic function we can think of. We will use this one as a model by which to compare all other quadratic functions we will examine.

VERTEX OF A PARABOLA All parabolas have a VERTEX, the lowest or highest point on the graph (depending upon whether it opens up or down.)

AXIS OF SYMMETRY All parabolas have an AXIS OF SYMMETRY, an imaginary line which goes through the vertex and about which the parabola is symmetric.

HOW PARABOLAS DIFFER Some parabolas open up and some open down.
Parabolas will all have a different vertex and a different axis of symmetry. Some parabolas will be wide and some will be narrow.

y = (x - 3)2 - 4 Vertex x y 6 5 5 4 -3 3 -4 2 -3 1 5 Example
Y-intercept x y Axis of Symmetry 6 5 5 Roots or x intercepts 4 -3 3 -4 Vertex 2 -3 1 5

The general form of a quadratic function is: f(x) = ax2 + bx + c The position, width, and orientation of a particular parabola will depend upon the values of a, b, and c.

Compare f(x) = x2 to the following: f(x) = 2x2 f(x) = .5x f(x) = -.5x2 If a > 0, then the parabola opens up If a < 0, then the parabola opens down

Now compare f(x) = x2 to the following: f(x) = x f(x) = x 2 - 2 Vertical shift up Vertical shift down

Now compare f(x) = x2 to the following: f(x) = (x + 2)2 f(x) = (x – 3)2 Horizontal shift to the left Horizontal shift to the right

When the general form of a quadratic function f(x) = ax2 + bx + c is changed to the vertex form: f(x) = a(x - h) 2 + k We can tell by horizontal and vertical shifting of the parabola where the vertex will be. The parabola will be shifted h units horizontally and k units vertically.

Thus, a quadratic function written in the form f(x) = a(x - h) 2 + k will have a vertex at the point (h,k). The value of “a” will determine whether the parabola opens up or down (positive or negative) and whether the parabola is narrow or wide.

f(x) = a(x - h) 2 + k Vertex (highest or lowest point): (h,k) If a > 0, then the parabola opens up If a < 0, then the parabola opens down

Axis of Symmetry The vertical line about which the graph of a quadratic function is symmetric. x = h where h is the x-coordinate of the vertex.

So, if we want to examine the characteristics of the graph of a quadratic function, our job is to transform the general form: f(x) = ax2 + bx + c into the vertex form: f(x) = a(x – h)2 + k

This will require to process of completing the square which is a little different than completing the square to solve a quadratic equation.

Graphing Quadratic Functions by Completing the Square

Factor x2 + 6x + 9 (x + 3)(x + 3) or (x + 3)2 Perfect Square Trinomial The factors are in the form (x + a)2 or (x - a)2. Note the relationship between the middle term and the last term. The last term is one-half the middle term squared. = 32 = 9 Find the value of the last term that will make the following perfect square trinomials. (x + 7)2 x2 + 14x + _______ 49 x2 + 7x + _______ x2 - 3x + _______

Changing from Standard Form to Vertex Form
Write y = x2 + 10x + 23 in the form y = a(x - h)2 + k. Sketch the graph. y = x x ( ) 1. Bracket the first two terms. y = (x2 + 10x + ____ - ____) + 23 25 25 2. Add a value within the brackets to make a perfect square trinomial. Whatever you add must be subtracted to keep the value of the function the same. y = (x2 + 10x + 25) y = (x + 5)2 - 2 3. Group the perfect square trinomial. (-5, -2) 4. Factor the trinomial and simplify.

Changing from Standard Form to Vertex Form
Write y = 2x2 - 12x -11 in the form y = a(x - h)2 + k. Sketch the graph. y = 2x x ( ) 1. Bracket the first two terms. y = 2(x2 - 6x) 2. Factor out the coefficient of the x2- term. y = 2(x2 - 6x + ____ - ____) - 11 9 9 y = 2(x2 - 6x + 9) 3. Add a value within the brackets to make a perfect square trinomial. Whatever you add must be subtracted to keep the value of the function the same. y = 2(x - 3)2 - 29 Multiply, when you remove this term from the brackets. 4. Group the perfect square trinomial. When grouping the trinomial, remember to distribute the coefficient. (3, -29) 5. Factor the trinomial and simplify.

Completing the Square y = -3x2 + 5x - 1 y = -3x2 + 5x - 1 ( ) y = -3(x x) - 1 y = -3(x x + ______ - ______ ) - 1 y = -3(x x ) y = -3(x ) y = -3(x )2 + Vertex is y = -3(x )2 +

The Vertex Formula

Using the general form, y = ax2 + bx + c, complete the square:
Completing the Square - The General Case Using the general form, y = ax2 + bx + c, complete the square: y = ax2 + bx + c y = (ax2 + bx ) + c The vertex is This IS the vertex BUT it is easier just to remember that the x-value is and then plug that in to the equation to get the y-value for the vertex.

Using the Vertex Formula
Find the vertex and the maximum or minimum value of f(x) = -4x2 - 12x + 5 using the axis of symmetry, the vertex is Find the x-value of the vertex: Find the y-value of the vertex: Therefore there is a maximum of y = 14, when x = The vertex is

Direction of the Parabola
If the coefficient of x2 is positive the parabola will open up. If the coefficient of x2 is negative the parabola will open down.

CHARACTERISTICS OF THE GRAPH OF A QUADRATIC FUNCTION
f(x) = ax2 + bx + c Parabola opens up and has a minimum value if a > 0. Parabola opens down and has a maximum value if a < 0.

EXAMPLE Determine without graphing whether the given quadratic function has a maximum or minimum value and then find the value. Verify by graphing. f(x) = 4x2 - 8x + 3 g(x) = -2x2 + 8x + 3

THE X AND Y INTERCEPTS OF A QUADRATIC FUNCTION
Find the x-intercepts by setting the quadratic function equal to zero and solve by whatever method is easiest. If the discriminant b2 – 4ac > 0, the graph of f(x) = ax2+ bx + c has two distinct x-intercepts and will cross the x-axis twice. 3. If the discriminant b2 – 4ac = 0, the graph of f(x) = ax2 + bx + c has one x-intercept and touches the x-axis at its vertex. 4. If the discriminant b2 – 4ac < 0, the graph of f(x) = ax2 + bx + c has no x-intercept and will not cross or touch the x-axis. 5. Find the y-intercept by substituting x=0 into function.