2.5 Continued
√b2-4ac (-4) 2-4(4)(1)=0
** Therefore there is only one POI between these two functions
Now try:
Find the point(s) of intersection for each pair of relations, and state whether the line is secant, tangent, or neither, to the curve.
A) y=2x2-5x+20 and y=6x-1 B) y=5x2+x-2 and y=-3x-6
C) The revenue for a production company by a theatre group is R(t)=50t2+300t and the cost for the production is C(t)=600-50t, where t is the ticket price in dollars.
Determine the ticket price if the production just breaks even. (Hint: Tangent)
By Celine Stoica, Kieran VanDalen, Jacob Lawrence, Taylor Mertens, and Sarah Durcikova
Quadratic Functions and
Equations
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2.1 Simplifying Radical Numbers
A Radical Number or an Irrational Number, is when x is not a perfect square, it is in radical form and it is the exact value of a number.
√a x √b = √ab
Simplifying Radicals
√50 √48÷√6
= √2 x √25 = √8
= √2 x 5 = √2 x √4
= 5 √2 = 2 √2
Multiplying Radicals
5√4 x 11√7 = 55√28 = 55√4√7 =55(2)√7 =110√7
**The 5 multiples with the 11, while the square roots multiple with each other. Then simplify √28**
Simplifying Radical Expressions
(6-√45)÷3 = (6-√9√5)÷3 = (6-3√5)÷3
= 3(2-√5)÷3 **Factor the 3 by the division of 3** = 2-√5
2.5 Continued
x2+2x-3=0 a=1 b=2 c=-3
=-2±√ 22-4(1)(-3) 2 =-2±√ 4+12 2
=-2+√16 and -2-√ 16 2 2 x=1 and x=-3
Sub 1 in y=2x+6 Sub -3 in y=2x+6 =2(1)+6 =2(-3)+6 y=8 y=0
~ These functions intersect at (1,8) and (-3,0) and it is secant
Ex 4: Determine the number of points of intersection of y=4x2+x-3 and y=5x-4
WITHOUT solving. Hint: use the discriminant
y=4x2+x-3 y=5x-4 (discriminant)
2.5 Continued
At the intersection y1=y2 4x+b=0.5x2+2x-8
-0.5x2-2x+8+b=0 -0.5x2+2x+8+b=0 a=-0.5x2 b=2x c=8+b
* No quadratic *
b2-4ac=0 discriminant formula (22)-(4)(-0.5)(8+b)=0
4+2(8+b)+0 4+16+2b+0 20+2b=0 2b=-20 2 2 b=-10
~ The y-intercept of the line is -10
Ex 3: Find the point(s) of intersection of the relation, state whether the line is secant, tangent, or neither, to the curve.
y=x2+4x+3 and y=2x+6
x2+4x+3=2x+6 x2+4x-2x+3-6=0
2.2 Finding a Max/Min of a Quadratic
There are three ways to write a Quadratic.
1) Vertex Form: y= a (x-h)²+k *Vertex is h, k
*Sign in the front of a gives direction of opening
2) Standard Form: y= ax²+bx+c *Y-int is (0,c)
*”a” gives direction of opening
3) Factored Form: y= a (x-r)(x-s)
*Roots/Solutions/X-int/Zeros
*Zeros= (r,0), (S,0)
*”a” gives direction of opening
*If a>0 (positive)- parabola opens up– minimum *If a<0 (negative)- parabola opens down– maximum
Find the minimum value of the following by completing the square
Ex. Y= 4x²-24x+31
1) Factor out the coefficient in front of the x² term but only from the first two terms
= 4 (x²-6x)+31 (-6/2)²=9
2.2 Continued
Y= 4 (x²-6x+9-9)+31
3) Factor the first 3 terms in the bracket as a PST
Y= 4 (x-3)²+31
4) 4th term isn’t used. Bring the 4th term outside the bracket by mul-tiplying it by the number in front.
Y= 4(x-3)²-36+31 **(4)(-9)
5) Add together the left over numbers Y= 4(x-3)²-5
Y= a (x-h)²+k
The vertex is (3,-5) a minimum value since the “a” value is positive (opens up)
Ex. 2. Find the vertex of y= -0.3x²-2.4x+7.3 and determine if it is a min or a max
Y= -0.3x²-2.4x+7.3 1) = -0.3 (x²+8x)+7.3
2) = -0.3 (x²+8x+16-16)+7.3 3) = -0.3 (x+4)²+7.3
4) = -0.3 (x+4)²+4.8+7.3 5) = -0.3 (x+4)²+12.1 6) A= -0.3 h= -4 k= 12.1
2.5 Continued
= 4.56+√ 6.9696 or 4.56-√ 6.9696 0.96 0.96
x=7.5 and/or x=2 Sub into either equation
Sub x=7.5 into y=0.24x+7.2
=0.24(7.5)+7.2 y=9
Sub x=2 into y=0.24x+7.2 =0.24(2)+7.2
y=7.68 è round to 7.7
~The functions intersect at (7.5,9) and (2,7.7)
Ex 2: If a line with slope 4 has only one point of intersection with the quadratic function y=½ x2+2x-8 what is the y-intercept of that line? Model your linear equation using slope y-intercept form
What we know,
“line with slope 4” mx+b, m=4, y=4x+b
2.5 Solving Quadratic/Linear systems
In a Quadratic/Linear system, you solve for where the two systems will intersect.
How many points of intersection exist? A system can have 2 maxi-mum.
You can have 2 points of intersection, a secant.
You can have 1 point of intersection, a tangent. * The tangent “grazes” *
Ex 1: Given a quadratic function of y=0.48x^2+4.8x and a linear function of y=0.24x+1.2 find the point(s) where the two functions in-tersect
Step 1: set the two equations equal to each other.
y1=y2 -0.48x2+4.8x=0.24x+7.2 0.48x2-4.8x+0.24x+7.2=0 0.48x2-4.56x+7.2=0 a=0.48 b=-4.56 c=7.2
Substitute into –(b)±√ (b) 2-4(a)(c) 2(a)
= -(-4.56)±√ (-4.56) 2-4(0.48)(7.2) 2(0.48)
= 4.56±√ 20.7936-13.824 0.96
2.2 Continued
**Therefore the vertex is (-4, 12.1) and it is a maximum value be-cause –0.3 is a negative (“a” value is negative)
Ex 3. Find the vertex of y= 4x²-24x+31
1) Take the first two terms and set them to =0 (standard form) 0= 4x²-24x
2) 4x (x-6)=0 common factor
4x= 0 x-6=0 X= 0 X=6
3) Find the average of these solutions; add them and divide by 2;
this is the x-value of the vertex 0+6
2 = 3
4) Sub the x-value into the original solution to find the y value X=3
Y= 4(3)²-24(3)+31 = 4(9)-72+31 = -5
2.2b Max/Min Applications
**Recall: If (h,k) is the vertex of the parabola y=a(x-h)^2+k, then k is the Mini-mum if a>0 or k is the MaxiMini-mum if a<0**
Example 1– Revenue Problem
The student council plans to run the annual talent show to raise money for charity. they would like to improve upon their profits from the previous year. after conducting a survey, they found that if they increased the ticket price by $1, 20 less people will attend. What ticket price will maximize their revenue?
**Important information: Tickets were $11 last year, 400 people attended, last year revenue R=$11*400 so R=$4400**
Let x be the number of $1 ticket price increases
R=(quantity)*(price) R=(# of tickets sold)*($) R=(400-20x)(11+1x)
lose 20 people each $1 increase $11 last year
R=(400-20x)(11+1x)
Now weFOIL
R=4400+400x-220x-20x^2
R=-20x^2+180x+4400
Lets use Partial Factoring
-20x^2+180x=0
2.4 Continued
=
×
=
√
2/3×2
=
√
2/6
So, we multiply the expression by a factor of 1, (
√
2/
√
2,
√
23/
√
23) and a radical multiplied by itself, becomes a
ra-tionalized number (
√
3×
√
3=3)
Examples To Try
A) 4√3-2√27
B)
(2√3)(4+5√3)C)
(2√3-√5)(3√6-√2)D)
5/2√-√32.4 Working with radical numbers
The algebra we use for adding and subtracting radical numbers are
radicals.
For example: 3x+6x=9x 3√5+6√5=9√5 Adding Radicals
5√8+3√18 =5√2√4+3√9√2 =5(2) √2+3(3) √3 =10√2+9√2 =19√2
Multiplying radicals:
(2√3)(3√6) =6√18 =6√9√2 =6×3√2 18√2
Rationalizing denominators:
**Just like we wouldn’t leave 2/4 but simplify it to 1/2, we have to properly simplify fractions containing radical numbers. It is NOT proper form to leave a radical in the denominator, there-fore we must rationalize denominators.
For example:
2.2b Max/Min Applications Continued
-20x*(x-9)=0 / \
-20x=0 x-9=0 x=0 x=9
Find average
x=0+9/2 x=4.5
~ the x value of the maximum is 4.5, $11+$4.50=$15.50 per per-son will maximize the revenue
substitute back in to R=-20x^2+180x+4400
R=-20(4.5)^2+180(4.5)+4400 R=$4805
~they will earn $4805 with the new ticket price
Try:
Example 2– Physics Problem
Nick threw a ball off of a roof. The ball fell to the ground and the equation is: h= -5t^2+15t+9
A) Determine the Maximum height of the ball
B) How long does it take the ball to reach its maximum height? C) How high is the roof?
2.3 Solving Quadratic Equations
Solving quadratic equations
There are three main ways to solve quadratic equations: 1) Quadratic formula
2) Factoring
3) Completing the square
1) Quadratic formula
The quadratic formula is
i.e. To find x the values of a, b, and c must first be determined from the given equation.
x = -4x2+12x-9
x = ax2+bx+c
a = -4
b = 12
c = -9
2.3b Continued
= 5
**Therefore yes, after 5 s
C) 150 m
0= -5t²+50t-150 A= -5 B=50 C= -150
=-50±√50²-(4)(-5)(-150) 2(-5)
=-50±√2500-3000 -10
=-50±√-500 -10
**Cannot go on with a negative under the discriminant
**Therefore it does not reach 150 m
2.3b Continued
A) 45 m B) 125 m C) 150 m
A) 45 m 45= -5t²+50t
0= -5t²+50t-45 A= -5 B= 50 C= -45
=-50±√50²-(4)(-5)(-45) 2(-5)
=-50±√2500-900 -10
=-50±√1600 =1 and 9
**Therefore yes, after 1 s and 9 s
B) 125 m 0=-5t²+50t-125 A= -5 B= 50 C= -125
=-50±√50²-(4)(-5)(-125) 2(-5)
=-50±√0 -10
2.3 Solving Equations Continued
The determined values are then used in the quadratic formula to solve for x.
Therefore the roots are x = (there is only one solution).
2) Factoring
To factor the quadratic equation it must be set to equal zero.
2.3 Continued
Therefore the solutions are -2 and (there are two solutions).
3) Completing the square.
Once more the equation must equal zero. i.e. 9x2 = 25
9x2 – 25 = 0
Now to find the difference of squares.
(3x-5)(3x+5)
x = x =
Therefore the solutions are and (there are two solutions).
Not all parabolas have the same amount of roots or solutions. To determine the amount of solutions the discriminant is used.
2.3b Continued
A=1 B=12 C=-6.25
=-12±√144-(4)(1)(-6.25) 2(1)
=-12±√144-(-25) 2
=-12+
√169 2 =-12+132 =0.5
And =-12-13
2
=-12.5 **Not possible as you cannot have a negative border
Therefore x=0.5 m and the border will be 0.5 m for the garden
Example 2– Airborne Object
2.3b Word Problems
Example 1– Area
A rectangular lawn measures 7 m by 5 m. A uniform border of flow-ers is to be planted along two adjacent sides of the lawn, as shown in the diagram. If the flowers that have been purchased will cover an area of 6.25 m², how wide is the border?
Find the border of this garden
(5+x)(7+x)
=35+5x+7x+x²-35
=x²+12x
=6.25m² 0=x²+12x-6.25
**Use Quadratic Formula** 5 m
7 m
x
2.3 Continued
The discriminant is the part underneath the radical of the quadrat-ic formula.
If ≥ 0 then there are two solutions.
If = 0 then there is one solution.
If ≤ 0 then there are no solutions.
i.e. 2x+3x-2 = 0 a = 2
b = 3
c = -2
9- (-16) 25
