Transcript UNIT A - Mr. Santowski

UNIT A
PreCalculus Review
7/17/2015
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Unit Objectives
• 1. Review characteristics of fundamental
functions (R)
• 2. Review/Extend application of function
models (R/E)
• 3. Introduce new function concepts pertinent to
Calculus (N)
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A3. Quadratic Functions
Calculus - Mr. Santowski
7/17/2015
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Lesson Objectives
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1. Factor and solve quadratic expressions/eqns
2. Find max/min on quadratic fcns
3. Sketch quadratic fcns
4. Quadratic models in physics (motion),
business (profit, cost), etc for finding roots,
max/min, points
• 5. Using technology, graph & determine where
a quadratic function in/decreases and its
concavity as well as roots and max/mins
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Fast Five
• 1. Sketch a graph that increases on (-2,4) U (5,8) and decreases on
(4,5)
• 2. Solve 2x2 - 12x - 7 = 0
• 3. Sketch the graph of a decreasing exponential function
• 4. How many roots does Q(x) = 3x2 + 4x - 8 have?
• 5. Evaluate tan(π/4)
• 6. Find the roots and vertex of f(x) = -0.5(x - 2)2 + 8. Sketch a graph
that is positive and increasing at x = 2
• 7. Sketch the graph of f(x) from Q6.
• 8. Factor 6x2 + x - 15
• 9. At what values is f(x) = (x-4)/(x-3) equal to 0?
• 10. Simplify x/5 + 5/x
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Extending your Knowledge and
Skills
• Consider f(x) = kx2 + (k+1)x - (k+2)
• 1. Graph the parabolas where k = 1,2,3,4. Compare the
vertices and general shapes of these parabolas. What
happens as k increases? Justify your answer.
• 2. Notice that all of these parabolas cross the x-axis at two
points. Let Sk be the larger of the two solutions of 0 = kx2 +
(k+1)x - (k+2). Using your calculator, find the values of S1, S2,
S3, S4.
• 3. Predict the behaviour of Sk as k gets very large. Provide
some proof for your hypothesis.
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Extending your Knowledge and
Skills
• Sketch a graph of a graph that has the following
features:
• (i) f is a function whose rate of change increases
on the interval (-4,6)
• (ii) This function, f , has its rate of change
decreasing on (6,+∞ )
• (iii) it goes through the point (2,-1)
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(A) Factoring Quadratic
Expressions
• Factor x2 + 6x + 8
• Factor x2 + 3x - 18
• General Strategy is …..?
• Factor 81x2 - 49
• Factor 36x2 - 25
• General strategy is ….
• Factor 2x2 - 7x + 3
• Factor 8x2 - 10x + 3
• General Strategy is ... ?
• Factor x2 + 6x + 9
• Factor 25x2 - 60x + 36
• General strategy is … ?
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(B) Solving Quadratic Equations
• We have three algebraic ways of solving
quadratic equations:
 (a) by factoring
 (b) by completing the square
 (c) by quadratic formula
• Solve the following QE using the three methods:
 (a) 2x2 + 7x = 4
 (b) 2x2 - 6x + 4 = 0
 (c) 2x2 = -1 + 6x
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(B) Solving Quadratic Equations
• We can solve the same three equations
graphically
 (a) by looking for x-intercepts of ax2 + bx + c = 0
 (b) by looking for intersection points of ax2 = bx + c or
ax2 + bx = c
• Solve the following QE using the graphing
methods:
 (a) 2x2 + 7x = 4
 (b) 2x2 - 6x + 4 = 0
 (c) 2x2 = -1 + 6x
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(C) Finding Maximums and
Minimums of Quadratic Eqns
• To find the max/min of a parabola, we have three
algebraic methods we can use:
• (1) completing the square method
• (2) averaging the roots to find the axis of symmetry and
then evaluating the function at that x-value
• (3) finding the axis of symmetry and then evaluating the
function at that x-value
• Example:
• Find the maximum point of f(x) = -3x2 - 12x - 9
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(C) Finding Maximums and
Minimums of Quadratic Eqns
• Find the maximum point of f(x) = -3x2 - 12x - 9
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Method 1 (C/S)
f(x) = -3[x2 + 4x + 3]
f(x) = -3[x2 + 4x + 4 - 4 + 3]
f(x) = -3[(x + 2)2 - 1]
f(x) = -3(x + 2)2 + 3
• So the vertex is at (-2,3) and since the parabola opens
down, the vertex represents a maximum point
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(C) Finding Maximums and
Minimums of Quadratic Eqns
• Find the maximum point of f(x) = -3x2 - 12x - 9
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Method 2 (averaging roots)
F(x) = -3(x2 + 4x + 3)
F(x) = -3(x + 3)(x + 1)
So the roots of f(x) = -3(x + 3)(x +1) are x = -3 and x = -1
Since the axis of symmetry cuts the parabola into 2 identical halves,
the axis must be halfway between x = -3 and x = -1
• So the axis of symmetry is at x = -2
• The point is then f(-2) = -3(-2)2 - 12(-2) - 9 = 3
• So the vertex is at (-2,3) and since the parabola opens down, the
vertex represents a maximum point
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(C) Finding Maximums and
Minimums of Quadratic Eqns
• Find the maximum point of f(x) = -3x2 - 12x - 9
• Method 3 (A/S)
• The axis can be simply found using the x = -b/2a calculation
(WHY??? Prove it!!!)
• Thus, x = -(-12)/[2(-3)] = 12/-6 = -2
• Then f(-2) is once again = -3(-2)2 - 12(-2) - 9 = 3
• So the vertex is at (-2,3) and since the parabola opens down, the
vertex represents a maximum point
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(D) Sketching Quadratic
Functions
• To sketch a quadratic function by hand
without technology, you simply need to
find the following:
• (1) roots
• (2) vertex (or max/min)
• (3) y-intercept
• Then plot these points and make the
sketch
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(D) Sketching Quadratic
Functions
• Working with f(x) = -3x2 - 12x - 9
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We know that the vertex is at (-2,3)
We know the roots are at x = -3 and x = -1
We know that the y-intercept is at f(0) = -9
We kow the parabola opens down
• So ….. plot points and make the curve
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(D) Sketching Quadratic
Functions
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(E) Quadratic Functions and
Projectile Motion
• The height of a thrown (or falling) object is a function of
its time in flight (h(t))
• This relationship between the height of a falling object
and time is given by the quadratic function h(t) = -0.5gt2
+ v0t + h0 where g is acceleration due to gravity (which
equals either 32 ft/s2 or 9.8 m/s2), v0 is the vertical
velocity of the object at t=0 and h0 is the height of the
object at t=0, t is time in seconds and h(t) is height in
meters or feet
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(E) Quadratic Functions and
Projectile Motion
• Mr. S stands on top of the new ASD math office (a height of 30 m),
and throws a baseball with an initial vertical velocity of 25 m/s. (Use
g = 10m/s2)
• (i) Determine the equation relating the height of the ball to its time of
flight
• (ii) Algebraically, determine the (a) vertex, (b) the x-intercepts, (c)
the y-intercept and interpret their values in context
• (iii) Find the domain and range of the function in context
• (iv) Use a graphing calculator to graph this relation. Sketch and label
key points
• (v) Find the average rate of change of height between t = 1 sec and t
= 2 sec
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(E) Quadratic Functions and Projectile
Motion - Ex 1 - Answers
• h(t) = -0.5(10)t2 + 25t + 30
• h(t) = -5t2 + 25t + 30
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Vertex => complete the square
h(t) = -5(t2 - 5t + 6.25 - 6.25) + 30
h(t) = -5(t - 2.5)2 + 31.25 + 30
h(t) = -5(t - 2.5)2 + 61.25
Vertex is at (2.5, 61.25) (which represents the maximum height and
the time at which it occurs)
• Alternative method => find A/S (x = -b/2a) and then evaluate h(-b/2a)
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(E) Quadratic Functions and Projectile
Motion - Ex 1 - Answers
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x-intercepts => F or QF or C/S
h(t) = 0 = -5t2 + 25t + 25
h(t) = 0 = -5(t2 - 5t - 6)
h(t) = 0 = -5(t - 6)(t + 1)
T = 6 sec or t = -1 sec (not possible) which represents the time after release
at which the ball lands
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Use C/S: h(t) = 0 = -5(t - 2.5)2 + 61.25
- 61.25/-5 = (t - 2.5)2
+√(12.25) = t - 2.5
2.5 + 3.5 = 6, -1 = t
y - int => let t = 0
h(0) = -5(0)2 + 25(0) + 30 = 30 which represents the starting height of the
ball
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(E) Quadratic Functions and Projectile
Motion - Ex 1 - Answers
• Domain {t E R | 0 < t < 6}
• You could argue a longer domain if you
wanted to tell me that the ball simply sits
on the ground for the extra time beyond t =
6 sec.
• Range { h E R | 0 < h < 61.25} for obvious
reasons!!
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(E) Quadratic Functions and Projectile
Motion - Ex 1 - Answers
QuickTime™ and a
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(E) Quadratic Functions and Projectile
Motion - Ex 1 - Answers
• Average rate of change (slope of secant)
h h(t 2 )  h(t1 )

t
t 2  t1
h(2)  h(1)
m sec 
2 1
2
2
5(2)

25(2)

30

5(1)
 25(1)  30



m sec 
2 1
60  50
m sec 
 10.0
1
• So on average, the height has increased at an average rate of 10.3
m/s between t = 1 s and t = 2 s.
m sec 

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(E) Quadratic Functions - Costs,
Revenues, Profits
• A shoe store owner has found that her revenue from
selling x shoes (where x is in thousands and R(x) is in
tens of thousands) is modeled by the equation R(x) = -x2
+ 30x, while the cost of producing the shoes in her
factory is modeled by the equation C(x) = 5x + 100
(where x is in thousands and C(x) is in tens of
thousands)
• (a) Find the minimum break-even quantity
• (b) Find the maximum revenue
• (c) Find the maximum profit
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(E) Quadratic Functions - Costs,
Revenues, Profits
• (a) Find the minimum break-even quantity
• Break-even means that revenue is equal to costs
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-x2 + 30 x = 5x + 100
0 = x2 - 25x + 100
0 = (x - 20)(x - 5)
So the break even points are x = 20 and x = 5
Thus the business has a minimal break even quantity of
5,000 shoes
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(E) Quadratic Functions - Costs,
Revenues, Profits
• (b) Find the maximum revenue:
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The revenue function is R(x) = -x2 + 30x
To find the maximum, simply factor
R(x) = -x(x - 30)
Meaning that the roots are at x = 0 and x = 30
Thus the axis of symmetry is at x = 15 and the max
revenue is realized when 15,000 shoes are sold
• The maximum revenue is R(15) = -(15)2 + 30(15) which
is $2,250,000
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(E) Quadratic Functions - Costs,
Revenues, Profits
• (c) Find the maximum profit
• Recall that profit is revenue - cost
• So P(x) = (-x2 + 30x) - (5x - 100)
• P(x) = -x2 + 25x + 100
• So the A/S is at x = -b/2a (x = -25/-2 = -12.5) so the max
profit is P(12.5)
• P(12.5) = -(12.5)2 + 25(12.5) + 100 = 56.25
• So the maximum profit happens when 12,500 shoes are
sold and the max profit is $562,500
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(E) Quadratic Functions and
Optimization
• I own an apartment building which has
1500 units, for which I charge $600 rent
per month. At this rent, all units are
occupied. I know that for every $25 rent
increase, I will lose 30 tenants. What rent
will maximize my revenue? What will be
the maximum revenue? How many units
are now vacated?
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(E) Optimization - Rent Question
• Revenue = (# of units) x (rent per unit)
• So initially, R = (1500 units) x (600$/unit) = $900,000
• Now introduce changes => let r represent the number of rent
increments being introduced
• Thus, rent per unit is now 600 + 25r (i.e. if I make 2
increments, then rent is now 600 + 25(2) = $650
• Thus, the # of units is now 1500 - 30r (i.e. if I made 2
increments, then I would loose 30x2 rental units and would
have 1500 - 30(2) = 1440 units still occupied
• So R = (1500 - 30r) x (600 + 25r)
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(E) Optimization - Rent Question
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So R = (1500 - 30r)(600 + 25r)
To find the optimal revenue, we need to find the vertex
We have several algebraic options
(i) rewrite in standard form and use A/S = -b/2a and then R(-b/2a) ==>
(ans: R(r) =-750r2 + 19500r + 900,000 and A/S is at r = 13)
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(ii) rewrite in standard form, complete the square and then we have the
vertex ==> (ans: R(r) = -750(r - 13)2 + 1,026,750
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(iii) find 2 x-intercepts and use then to find A/S and then R(-b/2a)
(easiest method since the eqn is already in factored form)
(ans: x-intercepts are 1500 - 30r = 0 (r = 50) and 600 + 25r = 0 (r = -24)
so A/S = 0.5(50 + -24) = 13 as before
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(E) Optimization - Rent Question
• Now answer the questions:
• (Ii) Optimal rent ==> 600 + 25r becomes 600 + 25(13)
= $925
• (ii) the optimal revenue will then be $1,026,750
• (iii) so the number of units vacated would be 1500 30(13) = 1110 still occupied, meaning that I have lost
30(13) units (390)
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(E) Optimization - Areas and Perimeters
• I want to build a fence around a rectangular field in my
backyard. I have 500 meters of fencing available and I will only
need to fence in 3 sides (see diagram). What dimensions on the
field will maximize the area of the field being fenced?
•
• W
L
W
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(E) Optimization - Areas and Perimeters
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So we need two equations:
(i) perimeter = 500 = 2 x width + length ==> So 500 = 2W + L
(ii) area = L x W
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We will work with the area equation since we want to optimize the area
But our area equation has 2 unknowns, so we will make a substitution
using the info in equation (i) ==> L = 500 - 2W
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Therefore, A = L x W
Which is now A = (500 - 2W) x W
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A(W) = -2W2 + 500W is our equation
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(E) Optimization - Areas and Perimeters
• So if A(W) = -2W2 + 500W, we now find the vertex again
using any algebraic method (completing the square,
finding A/S, or finding x-intercepts)
• Ans (ii) A(W) = -2(W - 125)2 + 31250 so vertex is at (125,
31250)
• Ans (ii) A/S = -500/(2 x -2) => W = 125 and A(125) =
31,250
• Ans (iii) x-intecepts of A(W) = (W)(500 - 2W) are W = 0
and W = 250, so A/S is at W = 125
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(F) Quadratics: Intervals of
Increase/Decrease & Concavity
• A fcn is increasing if for x2 > x1,
then f(x2) > f(x1)
• So the parabola illustrates an
interval of increase on (-2,∞ )
• A fcn is decreasing if for x2 >
x1, then f(x2) < f(x1)
• So the parabola illustrates an
interval of decrease on (-∞ ,-2)
• Functions may have points at
which the function neither
increases or decreases:
stationary points, extrema
QuickTime™ and a
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(F) Quadratics: Intervals of
Increase/Decrease & Concavity
• Ex 1: Given the parabolas on
the right, state the intervals of
increase and decrease
• Ex 2: Given the equations of
the following parabolas,
determine the intervals of
increase and decrease
• F(x) = -(x + 3)2 + 1
• G(x) = 2x2 + x - 10
• H(x) = (x + 4)(2 - 3x)
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(G) Quadratics - Concavity
• Simple functional definition of concave up
(we will see an alternate definition of
concavity later on): the graph bends
upward; the curve “opens” upward
• Simple functional definition of concave
down: the graph bends downward; the
curve opens downward
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(G) Quadratics - Concavity
• Ex: Sketch a graph that is:
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(1) concave up and decreasing on xER
(2) concave up and increasing on xER
(3) concave down and increasing on xER
(4) concave down and decreasing on xER
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(G) Quadratics - Concavity
QuickTime™ and a
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are needed to see this picture.
QuickTime™ and a
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QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
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(G) Quadratics & Concavity
• We can also see
concavity when
discussing quadratics:
see examples on the
side
• Ex. Is the function f(x)
= -(x + 4)^2 - 2
concave up or down?
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(G) Quadratics and Concavity
• An alternate definition of
concavity:
• If f is a function whose rate of
change increases as we move
from left to right, then the
graph of f is concave up
• If f is a function whose rate of
change decreases as we move
from left to right, then the
graph of f is concave down
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QuickTime™ and a
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(F) Internet Links
• From WTAMU:
• 1. Solving Quadratic Equations
• 2. Graphing Quadratic Functions
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From Purplemath
1. Factoring Quadratics
2. Graphing Quadratic Functions
3. Solving Quadratic Equations
4. Solving Word Problems
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(G) Homework
• Section 2.2, page 72-76
• (1) Q13,15,17,19 (work algebraically)
• (2) Q43,48,50,51,52,56,60,61
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