+ All Categories
Home > Documents > Graphing y = ax^2 + bx + c

Graphing y = ax^2 + bx + c

Date post: 30-Jun-2015
Category:
Upload: daisylistening
View: 984 times
Download: 0 times
Share this document with a friend
22
Graphing y = ax 2 + bx + c By L.D.
Transcript
Page 1: Graphing  y = ax^2 + bx + c

Graphing y = ax2 + bx + c

By L.D.

Page 2: Graphing  y = ax^2 + bx + c

Table of Contents Slide 3: Formula

Slide 4: Summary

Slide 5: How to Find the the Direction the Graph Opens Towards

Slide 6: How to Find the y Intercept

Slide 7: How to Find the Vertex

Slide 8: How to Find the Axis of Symmetry

Slide 9: Problem 1

Slide 16: Problem 2

Slide 22: End

Page 3: Graphing  y = ax^2 + bx + c

Formulay = ax2 + bx + c

Page 4: Graphing  y = ax^2 + bx + c

SummaryIn this presentation we are learning how to graph y = ax2 + bx + c. We will graph this by first finding the direction it opens up, the y intercept, the vertex and the axis of symmetry. The next three slides are devoted to how to find these.

Page 5: Graphing  y = ax^2 + bx + c

How to Find the the Direction the Graph Opens Towardsy = ax2 + bx + c

Our graph is a parabola so it will look like or

In our formula y = ax2 + bx + c, if the a stands for a number over 0 (positive number) then the parabola opens upward, if it stands for a number under 0 (negative number) then it opens downward.

Page 6: Graphing  y = ax^2 + bx + c

How to Find the y Intercept y = ax2 + bx + c

The y intercept is a number that is not generally used as a vertex, it is used as one of the places to plot the line. It’s formula is (0, c). The c is always a constant. The exception to it not being used as a vertex is when the b is equal to 0.

Page 7: Graphing  y = ax^2 + bx + c

How to Find the Vertexy = ax2 + bx + c

The vertex has an x coordinate of –b/2a

To find the y coordinate one must place the x coordinate number into the places x occupies in the problem.

Page 8: Graphing  y = ax^2 + bx + c

How to Find the Axis of Symmetryy = ax2 + bx + c

The line for the axis of symmetry crosses over the number achieved by doing the formula –b/2a.

Page 9: Graphing  y = ax^2 + bx + c

Problem 1Formula: y = ax2 + bx + c

y = 5x2 + 10x – 3

Directions: find the vertex, y-intercept and axis of symmetry. Then you may graph.

Page 10: Graphing  y = ax^2 + bx + c

Problem 1Formula: y = ax2 + bx + c

y = 5x2 + 10x – 3

The first thing we will find is the vertex. As mentioned in slide 6, this is done by first finding the x coordinate using –b/2a.

–b/2a = -10/2(5) = -10/10 = -1

Our x coordinate is -1. On the next slide we will find the y coordinate.

Page 11: Graphing  y = ax^2 + bx + c

Problem 1 Formula: y = ax2 + bx + c

y = 5x2 + 10x – 3

x coordinate: -1

As mentioned in slide 6, the y coordinate is found by placing the x coordinate in the places that x occupies in the problem.

y = 5(-1)2 + 10(-1) – 3

y = 5 + - 10 – 3

y = -8, so our y coordinate is -8, making our vertex located at (-1, -8).

Page 12: Graphing  y = ax^2 + bx + c

Problem 1Formula: y = ax2 + bx + c

y = 5x2 + 10x – 3

Vertex: (-1, -8)

Now we need to find the axis of symmetry, to do this we would use the same formula (–b/2a) as we used to get our x coordinate, so our axis of symmetry is -1.

Page 13: Graphing  y = ax^2 + bx + c

Problem 1Formula: y = ax2 + bx + c

y = 5x2 + 10x – 3

Vertex: (-1, -8)

Axis of symmetry: -1

The last step before graphing is where we need to find our y-intercept which will be the place that our vertex reaches too. We will do this by going to slide 6. The formula it gives us is (0, c), so our y-intercept is (0, -3).

Page 14: Graphing  y = ax^2 + bx + c

Problem 1Vertex: (-1, -8) (green)

Axis of symmetry: -1 (blue)

y-intercept: (0, -3) (red)

Now that we have all the information that is above gathered, we can safely graph. The colors that are in parenthesis are the colors the dots or lines will be.

Hint: The y intercept will be mirrored exactly due to the need of symmetry.

Page 15: Graphing  y = ax^2 + bx + c
Page 16: Graphing  y = ax^2 + bx + c

Problem 2Formula: y = ax2 + bx + c

y = x2 + 4x + 8

Directions: find the vertex, y-intercept and axis of symmetry. Then you may graph.

Page 17: Graphing  y = ax^2 + bx + c

Problem 2 Formula: y = ax2 + bx + c

y = x2 + 4x + 8

First we will find the vertex’s x-coordinate using –b/2a.

–b/2a = -4/2(1) = -4/2 = -2.

Since -2 is our x-coordinate we will now endeavor to find our y-coordinate.

y = (-2)2 + 4(-2) + 8

y = 4 – 8 + 8

y = 4, so our vertex is at (-2, 4)

Page 18: Graphing  y = ax^2 + bx + c

Problem 2Formula: y = ax2 + bx + c

y = x2 + 4x + 8

Vertex: (-2, 4)

Now we must find the axis of symmetry which is simply our x coordinate, -2.

Page 19: Graphing  y = ax^2 + bx + c

Problem 2Formula: y = ax2 + bx + c

y = x2 + 4x + 8

Vertex: (-2, 4)

Axis of symmetry: -2

We lastly need to find our y-intercept, which is (0, 8) when we follow our formula.

Page 20: Graphing  y = ax^2 + bx + c

Problem 2Vertex: (-2, 4) (green)

Axis of symmetry: -2 (blue)

y-intercept: (0, 8) (red)

Page 21: Graphing  y = ax^2 + bx + c
Page 22: Graphing  y = ax^2 + bx + c

Come to

myrata

temyhomework

.blogsp

ot.com to

ask q

uestions a

nd find m

ore m

ath

slidesh

ows.


Recommended