College Algebra
Answer/Discussion to Practice Problems
Tutorial 34:Graphs of Quadratic Functions
Answer/Discussion
to 1a


*Standard form of quad. function 
Since (h, k)
is the vertex in standard form, what do you think our vertex is for this
problem?
If you said ( 4, 2) you are correct.
Be careful about your signs on this problem. Notice how the sign
in front of h is a minus, but the one in front
of k is positive. So h is the number we are subtracting from x, which
in our case is negative 4. k is the number
we are adding at the end, which our case we are adding a negative 2. 
Maximum or Minimum?
Next we want to determine if the vertex that
we found, ( 4, 2), is a maximum or minimum point, without graphing.
If we know which direction the curve opens, that
can help us answer this question.
Since a = 1, and
1 is less than 0, this parabola would open down .
So does that mean the vertex is a maximum or minimum
point?
If you said a maximum point, you are right on.
So our vertex ( 4, 2) is the maximum point. 
Answer/Discussion
to 1b


*Identify a, b,
and c
*Plug values into vertex form. for a, b,
and c
*Plug 1 in for x to find the y value of the vertex 
The vertex would be (1, 1). 
Maximum or Minimum?
Next we want to determine if the vertex that
we found, (1, 1) , is a maximum or minimum point, without graphing.
If we know which direction the curve opens, that
can help us answer this question.
Since a = 1, and 1
is greater than 0, this parabola would open up .
So does that mean the vertex is a maximum or minimum
point?
If you said a minimum point, you are right on.
So our vertex (1, 1) is the minimum point. 
Answer/Discussion
to 2a

Since a = 1 and 1 > 0, then it looks like it
is going to curve up.
This gives us a good reference to know we are going in the right direction. 

*Standard form of quad. function

Since (h, k)
is the vertex in standard form, what do you think our vertex is?
If you said (2, 1) you are correct.
Be careful about your signs on this problem. Notice how
the sign in front of h is a minus, but the
one in front of k is positive. So h is
the number we are subtracting from x, which
in our case is 2. k is the number we
are adding at the end, which our case we are adding a 1. 
yintercept
Reminder that the yintercept is always
where the graph crosses the yaxis which means x = 0: 
The yintercept is (0, 5).
xintercept
Reminder that the xintercept is always
where the graph crosses the xaxis which means y = 0: 

*Replace y (or
f(x)) with 0

Note that this does not factor. Let's try solving by using
the quadratic formula: 

*Plug in values for a, b,
and c

Note how we got a negative number underneath the square root.
That means there is no real number solution. That also means that
there are NO xintercepts. 
Axis of symmetry
As shown on the graph, the axis of symmetry is x = 2. 
Answer/Discussion
to 2b

Since a = 1 and 1 < 0, then it looks
like it is going to curve down.
This gives us a good reference to know we are going in the right direction. 

*Identify a, b,
and c
*Plug values into vertex form. for a, b,
and c
*Plug 0 in for x to find the y value of the vertex 
yintercept
Reminder that the yintercept is always
where the graph crosses the yaxis which means x = 0: 
The yintercept is (0, 1).
xintercept
Reminder that the xintercept is always
where the graph crosses the xaxis which means y = 0: 
The xintercepts are (1, 0) and (1,
0). 
Axis of symmetry
As shown on the graph, the axis of symmetry is x = 0. 
Last revised on July 10, 2010 by Kim Seward.
All contents copyright (C) 2002  2010, WTAMU and Kim Seward. All rights reserved.

