april 9, 2015

Graphing Quadratic FunctionsToday:

NotebooksWarm-Up

Review: Vertex and Axis of Symmetry:The effects of a, b, & c on the parabola

Graphing Various Quadratic FunctionsClass Work: 4.1 (Front and back)

Graphing Quadratic Functions

x = 0

x = 1

(0, 2)

1. y = 4x2 – 72. y = x2 – 3x + 1

Find the axis of symmetry.

3. y = –2x2 + 4x + 3

(2, -12) 5. y = x2 + 4x + 5 6. y = -2x2 + 2x – 8

Find the vertex and state whether the graph opens up or down.

x = 𝟑

𝟐

Warm-Up

Graphing Quadratic FunctionsFor a quadratic function in the form y = ax2 + bx + c, when x = 0, y = c. The y-intercept of a quadratic function is c

Finding the Y intercept

Find the vertex and the y-intercept

1. y = x2 – 2 y = x2 – 4x + 4 y = -2x2 – 6x - 3

Graphing Quadratic FunctionsEffects of the a, b, & c values

With your graph paper, graph the function: y = x2

This is called the parent function. All other quadratic functions are simply transformations of the parent.

For the parent function f(x) = x2:• The axis of symmetry is x = 0, or

the y-axis.• The vertex is (0, 0)• The function has only one zero,

0.


The value of a in a quadratic function determines not only the direction a parabola opens, but also the width of the parabola.

Effects of the a, b, & c values


Example 1A: Comparing Widths of Parabolas Order the functions from narrowest graph to widest.

f(x) = 3x2, g(x) = 0.5x2, h(x) = 1.5x2

f(x) = 3x2

h(x) = 1.5x2

g(x) = 0.5x2

The function with the narrowest graph has the

greatest |a|.


The value of 'c' in a quadratic function determines not only the value of the y-intercept but also a vertical translation of the graph of f(x) = ax2 up or down the y-axis.

Tomorrow we look at how the 'b' value affects the parabola

Graphing Quadratic FunctionsComparing Graphs of Quadratic Functions

Compare the graph of the function with the graph of f(x) = x2

opens downward and the graph of

f(x) = x2 opens upward.

g(x) = −𝟏

𝟒x2 + 3• The graph of

Graphing Quadratic FunctionsCompare the graph of each the graph of f(x) = x2.g(x) = –x2 – 4

• The graph of g(x) = –x2 – 4 opens downward and the graph of f(x) = x2opens upward.

The vertex of g(x) = –x2 – 4 f(x) = x2 is (0, 0).

is translated 4 units down to (0, –4).

• The vertex of • The axis of symmetry is the same.


SOLUTIONIdentify the coefficients of the function. STEP 1

STEP 2 Find the AOS and the vertex. Calculate the x - coordinate.

Then find the y - coordinate of the vertex.

(–2) 2(1)= = 1x = b

2a –

y = 12 – 2 • 1 + 1 = – 2

The coefficients are a = 1, b = – 2, and c = – 1. Because a > 0, the parabola opens up.

Graph a function of the form y = ax2 + bx + c

y = x2 – 2x – 1Label the axis of symmetry. and the vertexGraph the function

Graphing Quadratic FunctionsGraph a function of the form y = ax2 + bx + c

SOLUTIONIdentify the coefficients of the function. STEP 1

STEP 2 Find the AOS and the vertex. Calculate the x - coordinate.x = b

2a =(– 8) 2(2)– –

Then find the y - coordinate of the vertex.y = 2(2)2 – 8(2) + 6 = – 2

So, the vertex is (2, – 2). Plot this point.

The coefficients are a = 2, b = – 8, and c = 6. Because a > 0, the parabola opens up.

= 2

y = 2x2 – 8x + 6.Graph

Graphing Quadratic FunctionsSTEP 3 Draw the axis of symmetrySTEP 4 Identify the y - intercept c,

STEP 5 Find the roots by using one of the solution methods, (factoring, for now)

(x - 3)(2x - 2); the solutions are:

Plot the point (0, 6). Then reflect this point overthe axis of symmetry to plot another point, (4, 6).

Plot the solutionsx = 3, x = 1STEP 6 Draw a parabola through

the plotted points.

y = 2x2 – 8x + 6. factor how?y = 2x2 – 6x - 2x + 6 =

x = 2.


STEP 1 Identify the coefficients of the function. STEP 2 Find the vertex. Calculate the x - coordinate.STEP 3 Draw the axis of symmetrySTEP 4 Identify the y - intercept c, STEP 5 Find the roots by using one of the solution methods,

We are unable to find the roots with our knowledge for now, so we'll select another value of x and solve for y. The AOS is 1, so let's choose x = -1. Find the y coordinate.

The two other points are (–1, 10) and (–2, 25) STEP 6 Reflect this point over the AOS to plot another point.

STEP 7 Graph the parabola

Graph a function of the form y = ax2 + bx + cy = 3x2 – 6x + 1, Plot 5 points and draw the curveGraph


x = 1

(–1, 10)

(0, 1)

(1, –2)

(–2, 25)

Graphing Quadratic FunctionsGraph a function of the form y = ax2 + bx + c

Step 1: Find the axis of symmetry.

Use x = . Substitute 1 for a and –6 for b.

The axis of symmetry is x = 3.

= 3

y = x2 – 6x + 9 Rewrite in standard form.y + 6x = x2 + 9Graph the quadratic function

Graphing Quadratic FunctionsStep 2: Find the vertex.

Simplify.= 9 – 18 + 9

= 0

The vertex is (3, 0).

The x-coordinate of the vertex is 3. Substitute 3 for x.

The y-coordinate is 0.

y = x2 – 6x + 9

y = 32 – 6(3) + 9



Step 3: Find the y-intercept.y = x2 – 6x + 9

y = x2 – 6x + 9

The y-intercept is 9; the graph passes through (0, 9).

Identify c.


Graphing Quadratic FunctionsStep 4 Find two more points on the same side of the axis of

symmetry as the point containing the y- intercept.

Since the axis of symmetry is x = 3, choose x-values less than 3.

Let x = 2y = 1(2)2 – 6(2) + 9

= 4 – 12 + 9= 1

Let x = 1 y = 1(1)2 – 6(1) + 9

= 1 – 6 + 9= 4

Substitutex-coordinates.

Simplify.

Two other points are (2, 1) and (1, 4).



Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points.

Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve.

y = x2 – 6x + 9

x = 3

(3, 0)

(0, 9)

(2, 1)(1, 4)

(6, 9)

(5, 4)

(4, 1)

x = 3

(3, 0)


Class Work 4.1

Front & Back

Show all work for credit

april 9, 2015

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