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Running head: COLLEGE ALGEBRA 1
College Algebra
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Institution

COLLEGE ALGEBRA 2
Question 1
The equation of the function in the plot is j ( x )=−x2−1. That is, option d. The graph has a y-
intercept of j ( x )=−1. Additionally, the function decreases for all value of x denoting that
j ( x ) =−x2−1 .
Question 2
The parabola f ( x ) =2 x2−8 x +3
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,
d f ( x )
dx = d
dx ( 2 x2−8 x +3 ) =4 x−8=0
4 x−8=0 , 4 x=8 , x= 8
4 =2
At x=2 , f ( x ) =2(2)2−8 ( 2 ) +3=8−16+3=−5
The vertex is at x=2∧ y=−5

COLLEGE ALGEBRA 3
Question 3
Part a
f ( x )=4−(x −1)2
f ( x ) =4− ( x2−2 x+1 )=3−x2 +2 x
We have the vertex when, d f ( x )
dx =0.That is ,
d
dx ( 3−x2 +2 x )=−2 x +2=0
−2 x+ 2=0 , 2 x =2 , x= 2
2 =1
When x=1 , f ( x )=4− (1−1 )2=4
The vertex is at the point (1 , 4)
The y-intercept is f ( 0 )=4− ( 0−1 )2=4−1=3
The x-intercept is f ( x )=3−x2+ 2 x=−x2 +2 x+ 3=0
x=−2 ± √22 −4 (−1 ×3)¿ ¿
2(−1)=−2 ± √16
−2 =−2± 4
−2 =1±−2
x=1+ ( −2 ) =−1∨x=1−−2=3
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 is x=1
Part c
The domain of the function from the graph is −∞ ≤ x ≤+∞ while the range is −∞ ≤ f ( x ) ≤ 4
Question 4
f ( x ) =−0.8 x2 +2.4 x +6
Part a
At maximum height, d f ( x )
dx =0= d
dx (−0.8 x2+2.4 x+6)