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SOLVING QUADRATIC EQUATIONS BY GRAPHINGBY L.D.
TABLE OF CONTENTS
Slide 3: Standard Form
Slide 4: Graph x2 + 4x = 5
Slide 13: Mini Lesson
Slide 14: Graph x2 + 4x = -6
STANDARD FORM
ax2 + bx + c = 0
PROBLEM 1
Graph x2 + 4x = 5
PROBLEM 1
Graph x2 + 4x = 5
Standard Form: ax2 + bx + c = 0
The first thing we need to do is get the problem in standard form. Above, you can see that the standard form is equal to zero. To do that we will completely clear one side of the problem. I choose the right side.
x2 + 4x = 5
-5
x2 + 4x – 5 = 0
PROBLEM 1
Graph x2 + 4x – 5 = 0
Standard Form: ax2 + bx + c = 0
The next thing that we do, now that the problem at the top is in standard form is to graph it. We will act as if the zero is a y*. To do this we will graph using the formula y = ax2 + bx + c. To start we will have to find the vertex, y-intercept and the axis of symmetry. We will do this on the next slide.
*I suggest that you review my slideshow called “Graphing y = ax^2 + bx + c” on my blog. You will need to know the terms I will use next.
PROBLEM 1Graph 1x2 + 4x – 5 = 0
Standard Form: ax2 + bx + c = 0
Formula: y = ax2 + bx + c
Vertex: To find the x coordinate of the vertex we use –b/2a.
–b/2a = -(4)/2(1) = -4/2 = -2 -2 = x-coordinate
Now we need to place what we got for x in the x place of the problem and solve for y.
1(-2)2 + 4(-2) – 5 = y
4 – 8 – 5 = y
-9 = y
So that makes our vertex (-2, -9)
PROBLEM 1
Graph 1x2 + 4x – 5 = 0
Standard Form: ax2 + bx + c = 0
Formula: y = ax2 + bx + c
Vertex: (-2, -9)
Now we need to find our y-intercept and axis of symmetry. Our y-intercept is just the only constant in the problem placed in the y spot of a coordinate to make (0, -5) and our axis of symmetry is the x-coordinate of the vertex placed in the x spot of a coordinate. So it is
(-2, 0).
PROBLEM 1
Vertex: (-2, -9) (red)
Y- Intercept: (0, -5) (green)
Line of Symmetry:(-2, 0) (blue)
Lastly we need to graph it all! The colors are the colors of the dots/lines I will use.
PROBLEM 1
The next thing we have to do is find where the lines hit the x-axis.
PROBLEM 1
The set of numbers are (-5, 0) and (1, 0). We will now substitute these into the math problem on the next slide. This is to check and see if we graphed correctly.
(1, 0)(-5, 0)
PROBLEM 1Coordinates that touched the x-axis: (-5, 0) and (1, 0)
Problem: 1x2 + 4x – 5 = 0
We will do each one separately, placing the x axis numbers in the x places of the problem.
(-5, 0):
1(-5)2 + 4(-5) – 5 = 0
25 – 20 – 5 = 0
0 = 0
(1, 0):
1(1)2 + 4(1) – 5 = 0
1 + 4 – 5 = 0
0 = 0
The fact that both the numbers reached zero proves that the equation was graphed and made correctly. Problem 1 is now done.
MINI LESSON
If the problems are graphed and the ends
don’t reach the x-axis, then the problem is considered
to be a no solution.
PROBLEM 2
Graph x2 + 4x = -6
PROBLEM 2Graph x2 + 4x = -6
We first need to place it in standard form which can easily be done by moving the -6 to the other side, making the problem x2 + 4x + 6 = 0. When we do the easy math we already know that the y intercept is (0,6). We will find the vertex below. (remember that for these equations, the 0 is thought of as a y)
-b/2a = -4/2(1) = -4/2 = -2
(-2)2 + 4(-2) + 6 = y
4 – 8 + 6 = y
2 = y
These equations make our vertex into (-2, 2) and our axis of symmetry be (-2, 0).
PROBLEM 2
Vertex: (-2, 2) (red)
Line of symmetry: (-2, 0) (blue)
Y- intercept: (0,6) (green)
Now that we have this information, we need to graph
PROBLEM 1
Since the parabola graphed doesn't’t touch the x-axis, it is considered to be no solution.
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