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




 

checkAnswer/Discussion to 1a

problem 1a


 
Vertex
Note how this quadratic function is written in standard form.  That means we can find the vertex by lining it up with the general standard form and identify (h, k).

 
answer discussion1a

 

*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 example 2d.

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.


 


 

 

checkAnswer/Discussion to 1b

problem 1b


 
Vertex
Note how this quadratic function is written in the form quadratic.  That means we can find the vertex by using the formula vertex.

 
answer 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 example 1c.

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.


 


 

 

checkAnswer/Discussion to 2a

problem 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. 


 

 
Just like in problem 1a above, this quadratic function is written in standard form.  That means we can find the vertex by lining it up with the general standard form and identify (h, k).

 
answer 2a1
*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.


 

 
y-intercept
Reminder that the y-intercept is always where the graph crosses the y-axis which means x = 0:

 
answer 2a2

*Replace x with 0

 
The y-intercept is (0, 5).
 

x-intercept
Reminder that the x-intercept is always where the graph crosses the x-axis which means y = 0:


 
answer 2a3

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

 


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

 
answer 2a4

*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 x-intercepts.

 

 
answer 2a5

 
Axis of symmetry
As shown on the graph, the axis of symmetry is x = -2.

 


 

 

checkAnswer/Discussion to 2b

problem 2bg


 

 
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. 


 

 
Just like in problem 1b above,  this quadratic function is written in the form quadratic.  That means we can find the vertex by using the formula vertex.

 
answer 2b1

 
 

*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


 
So the vertex is (0, 1).

 

 
y-intercept
Reminder that the y-intercept is always where the graph crosses the y-axis which means x = 0:

 
answer 2b2

*Replace x with 0

 
The y-intercept is (0, 1).
 

x-intercept
Reminder that the x-intercept is always where the graph crosses the x-axis which means y = 0:


 
answer 2b3
*Replace y (or f(x)) with 0

*Solve the quadratic by factoring


 
The x-intercepts are (-1, 0) and (1, 0).

 

 
answer 2b4

 
Axis of symmetry
As shown on the graph, the axis of symmetry is x = 0.

 

 

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Last revised on July 10, 2010 by Kim Seward.
All contents copyright (C) 2002 - 2010, WTAMU and Kim Seward. All rights reserved.