5 3 the graphs of quadratic equations-x

Graphs of Quadratic Equations

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations.


The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b,


For example, for y = -2x + 3, we start at the point (0, 3).

the graph of y = -2x + 3

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b,


For example, for y = -2x + 3, we start at the point (0, 3).

The graphs of 1st degree equations Ax + By = C are straight lines hence they are called linear equations. If we have a linear function in the form of y = f(x) = mx + b, then we may eyeball the graph, starting from the y-intercept b, eyeball the tilt using the slope m.



For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1,




For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3)




(0, 3)

∆y = -2





(0, 3)

(1, 1)∆y = -2 ∆x = 1





(0, 3)

(1, 1)∆y = -2 ∆x = 1




For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3) and the line contains both points.

For example, for y = -2x + 3, we start at the point (0, 3). The slope m = -2/1 means we may set ∆y = -2 with ∆x = 1, i.e. another point is located 2 down and 1 to the right from (0, 3) and the line contains both points.

The graphs of 2nd degree equations y = ax2 + bx + c are not straight. They are curves. We will develop a method for eyeballing the graphs of y = ax2 + bx + c in this section.

(0, 3)

(1, 1)∆y = -2 ∆x = 1




Graphs of Quadratic EquationsWe start with an example of a graph gives the general shape of the graphs of 2nd (quadratic) degree equations.


Example A. Graph y = –x2


Make a tableExample A. Graph y = –x2

x -4 -3 -2 -1 0 1 2 3 4



y

x -4 -3 -2 -1 0 1 2 3 4



y-16-9-4-1 0-1-4-9-16



x -4 -3 -2 -1 0 1 2 3 4

y-16-9-4-1 0-1-4-9-16

The graphs of 2nd (quadratic) equations are called parabolas. Graphs of Quadratic Equations

The graphs of 2nd (quadratic) equations are called parabolas. Parabolas describe the paths of thrown objects (or the upside-down paths).




Properties of Parabolas:



Properties of Parabolas: • Parabolas are symmetric with respect to a center line



Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex, or the tip.



Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex, or the tip.

the vertex

the vertex



Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex.The vertex is the starting point for graphing a parabola, i.e. the graph of y = ax2 + bx + c, a 2nd degree function.

the vertex

the vertex



Properties of Parabolas: • Parabolas are symmetric with respect to a center line• The highest or lowest point of the parabola sits on the center line. This point is called the vertex.

Vertex Formula: The vertex of y = ax2 + bx + c is at x = . –b2a

The vertex is the starting point for graphing a parabola, i.e. the graph of y = ax2 + bx + c, a 2nd degree function.

the vertex

the vertex

Graphs of Quadratic EquationsOne way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.


Example B. Graph y = x2 – 4x – 12



Vertex: set x = = 2 –(–4) 2(1)



Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4



Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)

Make a table centered at x = 2. x y 0 1 2 3 4 Note the y values are symmetric

around the vertex just as the points.

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)


around the vertex just as the points. (If they are not, check your calculation.)

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)



(2, -16)

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)



(2, -16)(3, -15)

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)



(2, -16)

(4, -12)

(3, -15)

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)



(2, -16)

(4, -12)

(3, -15)(1, -15)

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)



(2, -16)

(0, -12) (4, -12)

(3, -15)(1, -15)

–12 –15 –16–15–12



Vertex: set x = = 2 –(–4) 2(1)

(2, -16)

(0, -12) (4, -12)

Make a table centered at x = 2.

Note the y values are symmetric around the vertex just as the points. (If they are not, check your calculation.)

(3, -15)(1, -15)

One way to graph a parabola is to make a table around the vertex so the points will be plotted symmetrically.

x y 0 1 2 3 4

–12 –15 –16–15–12

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions.

Graphs of Quadratic EquationsWhen graphing parabolas, we must also give the x-intercepts and the y-intercept. The y-intercept is (0, c) obtained by setting x = 0. The x-intercept is obtained by setting y = 0 and solve the equation 0 = ax2 + bx + c which may or may not have real number solutions. Hence there might not be any x-intercept.

The center line is determined by the vertex.


The center line is determined by the vertex. Suppose we know another point on the parabola,


The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola.


The center line is determined by the vertex. Suppose we know another point on the parabola, the reflection of the point across the center is also on the parabola. There is exactly one parabola that goes through these three points.


(2nd way) To graph a parabola y = ax2 + bx + c.Graphs of Quadratic Equations

(2nd way) To graph a parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.-b2a



1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.

-b2a



1. Set x = in the equation to find the vertex.2. Find another point, use the y-intercept (0, c) if feasible.3. Locate its reflection across the center line. These three

points form the tip of the parabola.

-b2a




points form the tip of the parabola. Trace the parabola.

-b2a




points form the tip of the parabola. Trace the parabola.4. Set y = 0 and solve to find the x intercept.

-b2a





-b2a


Example C. Graph y = –x2 + 2x + 15




-b2a


The vertex is at x = 1, y = 16Example C. Graph y = –x2 + 2x + 15




-b2a


The vertex is at x = 1, y = 16y-intercept is at (0, 15)





-b2a


The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)





-b2a


The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0





-b2a


The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0





-b2a


The vertex is at x = 1, y = 16y-intercept is at (0, 15) Plot its reflection (2, 15)Draw, set y = 0 to get x-int:–x2 + 2x + 15 = 0x2 – 2x – 15 = 0(x + 3)(x – 5) = 0x = –3, x = 5





-b2a



Example C. Graph y = –x2 + 2x + 15 (1, 16)




-b2a



Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15)




-b2a



Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)




-b2a



Example C. Graph y = –x2 + 2x + 15 (1, 16)(0, 15) (2, 15)

(-3, 0) (5, 0)

Finally, we make the observation that given y = ax2 + …,if a > 0, then the parabola opens upward.


Finally, we make the observation that given y = ax2 + …,if a > 0, then the parabola opens upward.


if a < 0, then the parabola opens downward.

Exercise A. 1. Practice drawing the following parabolas with paper and pencil. Visualize them as the paths of thrown objects. Make sure pay attention to the symmetry.


Exercise B. Graph the parabolas by taking a table around the vertex that reflect the symmetry. Find the x and y intercepts.


4. y = x2 – 4 5. y = –x2 + 42. y = –x2 3. y = x2

6. y = x2 + 4 7. y = –x2 – 48. y = x2 – 2x – 3 9. y = –x2 + 2x + 310. y = x2 + 2x – 3 11. y = –x2 – 2x + 312. y = x2 – 2x – 8 13. y = –x2 + 2x + 814. y = x2 + 2x – 8 15. y = –x2 – 2x + 816. a. y = x2 – 4x – 5 b. y = –x2 + 4x + 517. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 519. y = x2 + 4x – 21 20. y = x2 – 4x – 45 21. y = x2 – 6x – 27 22. y = – x2 – 6x + 27

Exercise C. Graph the following parabolas by plotting the vertex point, the y–intercept and its reflection. Find the x intercepts.


23. y = x2 – 2x – 3 24. y = –x2 + 2x + 3

25. y = x2 + 2x – 3 26. y = –x2 – 2x + 3

27. y = x2 – 2x – 8 28. y = –x2 + 2x + 8

29. y = x2 + 2x – 8 30. y = –x2 – 2x + 8

31. y = x2 + 4x – 21 32. y = x2 – 4x – 45

33. y = x2 – 6x – 27 34. y = – x2 – 6x + 27Exercise D. Graph the following parabolas by plotting the vertex point, the y–intercept and its reflection. Verify that there is no x intercepts (i.e. they have complex roots). 35. y = x2 – 2x + 8 36. y = –x2 + 2x – 537. y = x2 + 2x + 3 38. y = –x2 – 3x – 439. y = 2x2 + 3x + 4 40. y = x2 – 4x + 32

Graphs of Quadratic EquationsAnswers to odd problems.

5. y = –x2 + 4

(0,0)

(0, 4)

(2, 0)

7. y = –x2 – 4

(0, –4)

(2, 0)(–4 , 0)

(–1, 9)

(–2, 0)

15. y = –x2 – 2x + 8 17. a. y = x2 + 4x – 5 b. y = –x2 – 4x + 5

19. y = x2 + 4x – 21 21. y = x2 – 6x – 27

(1, 0)(–5, 0)

(–2 , –9)(0, –5)

(3, 0)(–7, 0)

(–2 , –25)(0, –21)

(1, 0)(–5, 0)

(–2 , 9)

(0, 5)

(9, 0)(–3, 0)

(0 , –27) (0, –21)

(3 , –36)

23. y = x2 – 2x – 3(3, 0)(–1, 0)

(–1 , –4)

(0, –3)

3. y = x2

5 3 the graphs of quadratic equations-x

Education