Real-World Applications

MAT-121: COLLEGE ALGEBRA
Written Assignment 4
2.5 points each
SECTION 5.1
Algebraic
For the following exercise, rewrite the quadratic function in standard form and give the vertex.
1. y=4 x2+ 5 x −7
For the following exercise, determine whether there is a minimum or maximum value to the
quadratic function. Find the value and the axis of symmetry.
For a function with minimum the x^2 coefficient is greater than zero and a function with
maximum the coefficient is less than zero. The coefficients can be expressed as
y = a x2 +bx+ c
so, a = 4, b = 5, and c = 7. Since a > 0, the function has a minimum.
To get the value at the symmetry, one need to find the first derivative of the function y and
equate it to zero.
y=4 x2+ 5 x −7
dy
dx =8 x+5 =0
x = −5
8
This is the line of symmetry and is along the x-axis.
2. y= 1
3 x2−4 x +9
For the following exercise, rewrite the quadratic function in standard form and give the vertex.
Determine whether there is a minimum or maximum value to the quadratic function. Find the
value and the axis of symmetry. Determine the domain and range of the quadratic function.
ANSWER
The standard form of equation is y = a ( x−h )2+ k, where (h, k) is the vertex of the quadratic
equation. We use completing square method to get the standard form of the equation.
3y = ( x2−12 x )+ 27
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3y= ( x2−12 x+ ( 12
2 )
2
) +27−( 12
2 )
2
3y = ( x−6 )2+27−36
3y=( x−6 )2−9
y = 1
3 ( x−6 )2−3
From the standard form the vertex of the equation is (6, -3).
3. f ( x)=−1
2 x2 +4 x−2
Answer
−2 y=x2−8 x +4
−2 y= ( x2−8 x ) + 4
−2 y= (x2−8 x+ ( 8
2 )2
)+4− ( 8
2 )2
−2 y= ( x−4 ) 2 + 4−16
y = −1
2 ( x−4 )2 +6
From the standard form the vertex of the equation is (4, 6).
For the following exercise, use the vertex (h, k) and a point on the graph (x, y) to find the general
form of the equation of the quadratic function.
4. (h , k )=(−1 ,3) ,
(x , y )=(−2 , 6)
Answer
The standard form of the equation is;
y = a ( x−h ) 2+ k
6 = a (−2− ( −1 ) ) 2 +3
6 = a ( −1 ) 2 +3
a = 3
Thus, the standard form of the quadratic equation is;
y = 3 ( x+ 1 )2+3
The general form is;
y = 3 ( x2+ 2 x +1 ) +3
y = 3 x2+ 6 x +6
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Real-World Applications
5. An athletic stadium holds 105,000 fans. With a ticket price of $22, the average
attendance has been 32,000. When the price dropped to $16, the average attendance rose
to 50,000. Assuming that attendance is linearly related to ticket price, what ticket price
would maximize revenue? Round ticket price to the nearest ten cents.
Answer
The points of the line are (22, 32000) and (16, 50000)
Gradient = m = 50000−32000
16−22 = -3000
A = −3000 t+ b
When ticket cost $16 the attendance is 50,000.
50,000 = -3,000 (16) + b
b = 98,000
The linear equation is
Q = -3000t + 98,000
The total revenue will be
= Price∗Quantity
= −3000 t2 + 98000t
The marginal revenue will be
= d
dt (total revenue)
= d
dt (−3,000 t2+ 98,000 t)
= −6,000 t + 98,000
The total revenue will be maximized when the marginal revenue is equal to zero.
-6,000t + 98,000 = 0
t = 98,000
6,000
t = 16.33
Thus, the ticket price should be at $16.33 to maximize revenue.
SECTION 5.2
Algebraic
For the following exercise, identify the function as a power function, a polynomial function, or
neither.
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6. f ( x)=4 ( x3 )
3
Answer
This can be expressed as
f ( x ) = 4 x9
Power function can be expressed as; y = k x p; where k and p are real numbers. Thus, the
function f ( x )is as power function.
For the following exercise, find the degree and leading coefficient for the given function if it is a
polynomial. If it is not a polynomial, then state so.
7. f ( x)= ( 2 x2−5 )
2
+ ( x−3 ) 2 +5
Answer
We write the function in a general form;
f ( x ) = 4 x4−20 x2+ 25+ ( x−3 )2 +5
= 4 x4−20 x2+ 25+ x2−6 x+ 9+5
= 4 x4−19 x2−6 x +39
The leading coefficient is; 4 and is of degree 4.
For the following exercise, find the intercepts of the functions.
8. g( x )= ( 3 x2−10 x−8 ) ( x +3 )
Answer
Expanding the function, we have;
g ( x ) = 3 x3−x2−38 x−24
The y-intercept is (0, -24)
This will have three x-intercepts, when g ( x ) = 0;
3 x3−x2−38 x−24=0
By factorization;
( 3 x2−10 x−8 ) ( x +3 )=0
( 3 x+ 2 ) ( x−4 ) ( x +3 ) =0; x = −2
3 , x= 4, x = -3
The intercepts are at the point (−2
3 , 0 ), ( 4 , 0 )∧ (−3 , 0 )
Technology
For the following exercise, graph the polynomial function using a calculator or a graphing utility.
Based on the graph, determine the intercepts and the end behavior.
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