
a
To graph: the
a

Explanation of Solution
Given information:
The quadratic equation is y=0.5x2+2x−2 .
Graph:
The graph mentioned below is in real number line of x axis and y axis.
Where vertex (−10,−60) .
Interpretation:
Consider the quadratic equation y=0.5x2+2x−2 .
The mentioned quadratic equation above will be in vertex form y=a(x−h)2+k where (h,k) is the vertex.
We can write the mentioned quadratic equation in the vertex form as y=0.5x2+2x−2⇒y=0.5(x2+20.5x−20.5)⇒y=0.5(x2+20x−20)⇒y=0.5(x2+2*x*10+102−102−20)⇒y=0.5((x+10)2−102−20)⇒y=0.5((x+10)2−120)⇒y=0.5(x+10)2−60.
Therefore the vertex form of the equation is y=0.5(x+10)2−60 .
Which is comparable to y=a(x−h)2+k where a=0.5 which is greater than 0 .
Since a>0 the direction of graph will be positive side of y axis and the vertex is (−10,−60) .
The discriminant of standard form of quadratic equation y=0.5x2+2x−2 is D=b2−4*a*c⇒D=(2)2−4*(0.5)*(2)=4−4=0. .
Here D=0 .
Therefore the graph will lie on x axis.
At x=0 the y=0.5x2+2x−2 will yield y=−2 .
Therefore graph will cut y axis at (0,−2) .
Therefore the graph will be as mentioned below.
b
To draw: the table for the equation y=0.5x2+2x−2 .
b

Answer to Problem 2E
The table of the equation is
xy−50.5−4−2−3−3.5−2−4−1−3.50−210.5 .
Explanation of Solution
Given information:
The equation is y=0.5x2+2x−2 and the value of x is −5,−4,−3,−2,−1,0,1.
Formula:
We will substitute the value of x in the equation y=0.5x2+2x−2 .
Calculation:
Consider the equation y=0.5x2+2x−2 .
Since the value of x is −5,−4,−3,−2,−1,0,1.
Recall that we will substitute the value of x in the equation y=0.5x2+2x−2 .
Therefore
At x=−5⇒y=0.5(−5*−5)+2(−5)−2=0.5 .
At x=−4⇒y=0.5(−4*−4)+2(−4)−2=−2 .
At x=−3⇒y=0.5(−3*−3)+2(−3)−2=−3.5 .
At x=−2⇒y=0.5(−2*−2)+2(−2)−2=−4 .
At x=−1⇒y=0.5(−1*−1)+2(−1)−2=−3.5 .
At x=0⇒y=0.5(0*0)+2(0)−2=−2
At x=1⇒y=0.5(−1*−1)+2(−1)−2=0.5.
Therefore the table will be as mentioned below.
xy−50.5−4−2−3−3.5−2−4−1−3.50−210.5
Chapter 4 Solutions
Glencoe Algebra 2 Student Edition C2014
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