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
6. 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.
7. PROBLEM 1
Graph 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)
8. 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).
9. 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.
10. The next thing
we have to do is
find where the
PROBLEM 1 lines hit the x-
axis.
11. The set of
numbers are
(-5, 0) and (1, 0).
PROBLEM 1 We will now
substitute these
into the math
problem on the
next slide. This
is to check and
see if we
(-5, 0) graphed
(1, 0)
correctly.
12. PROBLEM 1
Coordinates 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.
13. 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.
15. PROBLEM 2
Graph 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).
16. 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
17. Since the
parabola
graphed
PROBLEM 1 doesn't’t touch
the x-axis, it is
considered to
be no solution.
