College Algebra Tutorial 34


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



WTAMU > Virtual Math Lab > College Algebra > 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.
 

(return to problem 1a)


 

 

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.
 

(return to problem 1b)


 

 

checkAnswer/Discussion to 2a

problem 2a
 

Step 1: Does the graph curve up or down?

 
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. 
 

Step 2: Find the vertex.

 
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.
 

Step 3: Find the intercepts.

 
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

*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 x-intercepts.

 
Step 4: Graph the parabola.

 
answer 2a5

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

 
(return to problem 2a)


 

 

checkAnswer/Discussion to 2b

problem 2bg
 

Step 1: Does the graph curve up or down?

 
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. 
 

Step 2: Find the vertex.

 
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).

 
Step 3: Find the intercepts.

 
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).

 
Step 4: Graph the parabola.

 
answer 2b4

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

 
(return to problem 2b)

 

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WTAMU > Virtual Math Lab > College Algebra >Tutorial 34: Graphs of Quadratic Functions


Last revised on July 10, 2010 by Kim Seward.
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