College Algebra - Solved Problems and Examples

Running head: COLLEGE ALGEBRA 1
College Algebra
Name
Institution

COLLEGE ALGEBRA 2
Question 1
The equation of the function in the plot is g ( x )=(x +1)2+ 1. That is, option b. The graph is
obtained from f ( x )=x2 by an horizontal shift of 2 units to the left followed by a vertical shift of
1 unit up.
Question 2
The parabola f ( x ) =3 x2−12 x+ 1
The vertex of the parabola is the stationary minimum point. We get the point by getting the
derivative of the function f ( x ) and then equating it to zero. That is,
df ( x )
dx = d
dx ( 3 x2−12 x+ 1 )=6 x−12=0
6 x−12=0 ,6 x =12, x= 12
6 =2
At x=2 , f ( x ) =3(2)2−12 ( 2 ) +1=12−24+1=−11

COLLEGE ALGEBRA 3
The vertex is at x=2∧ y=−11. Therefore, the answer is (2,-11), option c.
Question 3
Part a
f ( x ) =x2−2 x−3
We have the vertex when, df ( x )
dx =0.That is ,
d
dx ( x2−2 x−3 ) =2 x−2=0
2 x−2=0 , 2 x=2 , x =2
2 =1
When x=1 , f ( x ) =12−2 ( 1 ) −3=−4
The vertex is at the point (1 ,−4)
The y-intercept is f ( 0 ) =02 −2 ( 0 ) −3=0−3=−3
The x-intercept is f ( x )=x2−2 x−3=0
x=2 ± √ (−2)2−4 (1 ×−3)¿ ¿
2× 1 =2 ± √ 16
2 =2 ± 4
2 =1± 2
x=1+2=3∨x=1−2=−1
Using the obtained vertex and intercepts, we can plot the graph of the function as shown in the
figure below.

COLLEGE ALGEBRA 4
Part b
The axis of symmetry of the parabola is x=1
Part c
The domain of the function from the graph is −∞ ≤ x ≤+∞ while the range is −4 ≤ f ( x ) ≤ ∞
Question 4
Let the two number be x and y.
∑ , x + y=16 , y =16−x
The product, P=xy =x ( 16−x ) =16 x−x2
The maximum product occurs when, dP
dx =0. That is,
dP
dx = d
dx ( 16 x−x2 )=16−2 x=0