Math 112 Quadratics- Notes
Note(30) Quadratic functions come in the form:
y = ax2 + bx + c where a ¹ 0
Very often a quadratic function will touch or cross the x- axis. The point(s) where the function crosses are significant as they often have special meaning. These points contain the zeroes or roots of the quadratic.
y = x2 +1 no x intercepts
no zeroes of the quadratic function
no Real roots, the roots are imaginary
y = x2 one x intercept
one zero of the quadratic function
one root of the quadratic equation
Math 112 Quadratics- Notes
y = x2 -3 two x intercepts
two zeroes of the quadratic function
two roots of the quadratic equation
Roots or zeroes
For the above graph the points on the x-axis are called intercepts.
The x values of these points are called:
1) the roots of the quadratic equation or
2) the zeroes of the quadratic function
Math 112 Quadratics- Notes
Note(31) Quadratic Equation- a quadratic equation comes in the form:
0 = ax2 + bx + c where a¹ 0 and y= 0
When the quadratic function has y set to 0 it is called a quadratic equation. Solving for the variable “x” gives us the values of the “roots” of the equation or “zeroes” of the function.
For the quadratic function y = x2 +2x –8, the roots can be found by solving the quadratic equation when y is set to “0”. We solve for variable in the equation: 0 = x2 +2x –8 and obtain the roots: x = 2 and x = -4. These values may be obtained a number of ways….
Method #1 Graphing calculator method
Enter the function after y1= x2 + 2x –8
Press 2nd trace option 2:zeroes. Set left boundary, enter, and right boundary, enter, then press enter to guess the value of the root.
Method #2 Factoring method
Set y = 0 then factor the quadratic equation
0 = x2 +2x –8
0 = (x-2) (x+4) set x-2 = 0 set x+4 = 0
x = 2 x = -4
The roots of the equation are 2 and –4
The submarine will surface in 2 hours. (-4 is an inadmissible time)
Method #3 completing the square method
0 = x2 +2x –8 Add 8 to both sides to move constant
8 = x2 + 2x Find the missing 3rd term of the trinomial
8 + 1 = x2 +2x + 1 (b/2)2 = (2/2)2 = (1)2 = 1
9 = (x+1)2 Take the square root of both sides
±3 = (x+1) Subtract 1 from both sides
-1 ± 3 = x
The roots are: x = -1 +3 and x = -1 – 3
x = +2 x = -4
The submarine will surface in 2 hours. (-4 is an inadmissible time)
Asn(C1) Q 8 a c d ,9 a b ,10 a) g) p. 45
Math 112 Quadratics- Notes
Method #4 using the quadratic formula
The quadratic formula obtained from completing the square using the general form of the quadratic y = ax2 +bx +c is:
x = -b ± Ö(b)2 – 4ac
2a
For the equation y = x2 +2x –8 the values of a, b, and c are a =1 b = 2 c= -8
Substituting into the formula we have:
x = - (2 ) ± Ö(2 )2 – 4( 1 )(-8 ) x = -2 + 6 = 4 = +2
2( 1 ) 2 2
x = -2 ±Ö4+32
2 x = -2 – 6 = -8 = -4
x = -2 ± Ö36 x = -2 ±6 2 2
2 2
The roots are +2 and –4
The submarine will surface in 2 hours. (-4 is an inadmissible time)
Asn(C2) Q 24 p. 48 Advanced: Derive the quadratic formula by hand
Math 112 Quadratics- Notes
Note(32) Using the quadratic formula to find the two roots of the equation gives us the x values of the intercepts. Averaging these values will give us the center point of the parabola along the x- axis through which the axis of symmetry runs. The vertex shares this same x-value. The average for the two x values expressed in general form is:
-b + Ö(b)2 – 4ac + -b - Ö(b)2 – 4ac
2a 2a = -b
2a
2
This expression (–b/2a) is the midpoint of the parabola along the x axis and therefore the x value in the vertex. We may find the vertex by solving for x then substituting this value into the quadratic function to find the y value. e.g. Find the vertex of the parabola y = x2 +2x – 8.
Midpoint = -b = -(2) = -2 = -1 Coordinate of midpoint (-1, 0)
2a 2(1) 2
Since the vertex is directly below/above the midpoint, the coordinate of the vertex is: (-1 , ?)
Now substitute –1 into the quadratic function to find the y value of the vertex.
y = x2 +2x -8
= (-1 )2 + 2(-1 ) -8
= 1 – 2 – 8
y = -9
The vertex has coordinates (-1,9)
Asn(C2) cont’d Q 25 p. 48
Math 112 Quadratics- Notes
Note(33) The quadratic formula will sometimes produce an answer that does not belong to the Real number system. In such a case the quadratic has no Real roots so that the graph has no x-intercepts. The roots are complex numbers and belong to the imaginary number system.
E.g. Find the roots of the equation 0 = x2 –6x +13
Use the quadratic formula:
x = -b ± Ö(b)2 – 4ac
2a
x = - (-6 ) ± Ö(-6 )2 – 4(1 )(+13 ) a= 1 b = -6 c = +13
2(1 )
x = +6 ± Ö 36 – 52
2
x = +6 ± Ö-16 Rewrite -16 as 16 ´ -1
2
x = +6 ± Ö 16 ´ -1
2
x = +6 ± Ö16 Ö-1 Separate 16 and –1 into two radicals
2
x = +6 ± 4 Ö-1 Take the square root of 16
2
x = +6 ± 4i Write Ö-1 as i, an imaginary number
2
x = +3 ± 2i Reduce by dividing each term by 2
The roots of the quadratic are:
X = +3 + 2i and x = +3 – 2i
Note: These are imaginary roots and therefore there are no x-intercepts.
Asn(C3) Q 27,28,29 p. 48,49
Asn(C4) Q 30 a,c,e,g 31 a,b p. 49 Advanced Q 32 a , c, e p. 49
Math 112 Quadratics- Notes
Note(34) Problem solving often requires that we solve for the vertex and/or the roots of the equation. The vertex gives the maximum or minimum value of a question whereas the roots give the “time to hit the ground” or the value when there is “no profit” or “width” when the area is zero etc.
Eg. a golf ball is hit from a 1 meter platform inside a golf dome and follows a path given by y = -6x2 +12x +1 where x = time (s) and y = height (m)
1) What maximum height will it reach? (Hint: find the vertex)
Vertex formula = -b = - (12) = -12 = +1 = vertex = (+1 , ?)
2a 2(-6) -12
To find the y value in the vertex we substitute +1 into the formula as follows:
y = -6x2 +12x + 1
y = -6(+1)2 +12(+1) +1
y = -6(+1) + 12(+1) +1
y = -6 +12 +1
y = +7
The vertex has coordinates (1,7) which we interpret as after 1 second the golf ball reaches a maximum of 7 meters.
2) When will it hit the ground? (Hint: It hits the ground at the x- intercept)
Find the zeroes or roots of the quadratic equation:
x = -b ± Ö(b)2 – 4ac
2a -12 ± 13 -12 +13 = 1 = -0.8
x = - (12 ) ± Ö(12 )2 – 4(-6 )(1) -12 -12 -12
2(-6)
x = -(12) ± Ö144 +24) -12 –13 = -25 = +2.08
-12 -12 -12
x = -12 ± Ö 168
-12
The golf ball will hit the ground after 2.08 seconds. The first answer is inadmissible as it is not sensible to have a negative answer.
Asn(C5) Q 22 p. 33, Q 26 p. 34, Q 5 p. 44, Q 39,40,41 p. 53
Asn(C6) Q 34,35,36,38 p. 52,53
Math 112 Quadratics- Notes
Note(35) The discriminant of the quadratic formula is given by the expression b2 – 4ac. The value of the expression can assist us in determining the nature of the roots for a quadratic equation.
e.g Find the discriminant for the following quadratic by using the expression b2 –4ac . Also find the roots of the equation.
Example #1 y = x2 –6x +13 a = 1 b = -6 c = 13
The value of the discriminant is:
b2 –4ac
(-6 )2 – 4(1 )(13)
36 – 52
-16 Note: The discriminant is negative
Find the roots:
x = -b ± Ö(b)2 – 4ac
2a
x = -(-6 ) ± Ö-16 (from above)
2(1)
x = +6 ± Ö16 ´ -1 -16 may be written as 16 ´ -1
2
x = +6 ± Ö16 Ö-1 Ö16 ´ -1 may be separated into Ö16 ´Ö-1
2
x = +6 ± 4i Take the square root of 16 and replace Ö-1 with i
2
x = +6 ± 4i Separate into two fractions and reduce
2 2
x = 3 ± 2i which gives the 2 roots…… 3 + 2i and 3 – 2i
Conclusion: When the discriminant is negative we obtain 2 complex roots. The graph obtained indicates that there are no x-intercepts:
Math 112 Quadratics- Notes
Example #2 y = x2 +4x + 4 a = 1 b = +4 c = 4
The value of the discriminant is:
b2 –4ac
(+4 )2 – 4(1 )(4)
16 - 16
0 Note: The discriminant is zero
Find the roots:
x = -b ± Ö(b)2 – 4ac
2a
x = - (4 ) ± Ö 0 (from above)
2(1)
x = -4
2
x = -2 One answer
By factoring we discover that there are two roots (not one) but they are identical. We can factor the trinomial to see the two roots:
0 = x2 +4x +4
0 = (x +2)(x + 2)
First factor: Second factor:
We set x +2 = 0 We set x +2 = 0
Solve: x = -2 Solve: x = -2
Conclusion: When the discriminant is zero we obtain 2 real roots that are identical. The graph obtained indicates that there is one x-intercept:
(-2 , 0)
Math 112 Quadratics- Notes
Example #3 y = x2 -x -6 a = 1 b = -1 c = -6
The value of the discriminant is:
b2 –4ac
(-1)2 – 4(1 )(-6)
1 +24
25 Note: The discriminant is positive
Find the roots:
x = -b ± Ö(b)2 – 4ac
2a
x = -(-1 ) ± Ö 25 (from above)
2(1)
x = +1 ± 5
2
x = 6 = +3 or x = -4 = -2 The roots are +3 and -2
2 2
Conclusion: When the discriminant is positive we obtain 2 real roots. The graph obtained indicates that there are two x-intercepts.
(0,-2) (0,+3)
Summary of the discriminant:
If b2 – 4ac < 0 The graph has two imaginary roots and no x-intercepts.
If b2 – 4ac = 0 The graph has two identical real roots and one x-intercept.
If b2 – 4ac > 0 The graph has two real roots and two x-intercepts.
Note: All quadratics have two roots. It is the x-intercepts that vary.
Asn(C7) Q 53,56,57, 58 a b, 59 p.56,57
Asn(C8) Handout on various problems involving quadratics. Selected.
Test #3
Asn(C9) Advanced Q 62,64,63 p. 57
Advanced Q Investigation #7 65,66,67 p. 58
Advanced Q 68 to 72 p. 59 selected questions.