#$&*
course Mth 158
7/17 145
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
030. * 30
*********************************************
Question: * 4.3.36 / 4.3.42 (4.1.42 7th edition) (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The function of x^2 - 2x - 3 has an a=1 b= -2 and c= -3.
The function will open upward due to the fact that a > 0.
We find the axis of symmetry by
-b / 2a = -(-2)/2*1 = 1
We find the vertex by plugging f(1) into the original equation:
1^2 -2(1)-3 giving us -4
So our vertex is (1, -4)
We find the x intercepts by x^2 -2x - 3=0
(x-3)(x+1)
(x-3)=0
x=3
(x+1)=0
x=-1
So the x intercepts are (3, 0) and (-1, 0)
The y intercepts occurs when x is equal to 0. Y=f(0) = -3
(0, -3)
With the vertex being (1, -4) our lowest possible y value will be -4. We assume that everything greater than -4 will appear giving us the range of y > -4.
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
* * The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3.
The graph of this quadratic function will open upwards, since a > 0.
The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4).
The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get
(x - 3) ( x + 1) = 0 so that
x - 3 = 0 OR x + 1 = 0, giving us
x = 3 OR x = -1.
So the x intercepts are (-1, 0) and (3, 0).
The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3).
The function can be evaluated for any real number x, so its domain is the set of all real numbers.
The graph is a parabola opening upward, with vertex (1, -4). So the lowest possible y value is -4, and all values greater than -4 will occur. So the range is y > -4.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique Rating: 3
*********************************************
Question: * 4.3.51 / 4.3.57. graph of parabola vertex (1, -3), point (3, 5)
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(y-k)=a(x-h)^2 with h=1 k= -3
y- (-3)= a(x-1)^2
Substitute (3, 5) for x and y
5-(-3)= a(3-1)^2
8=a4
a=2
I’m not sure where to go from here.
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
* * The graph is a parabola with vertex (1, -3), so it has form
(y - k) = a ( x - h)^2, with h = 1 and k = -3.
Thus we have
y - (-3) = a (x - 1)^2,
and the only unknown is a. To find a we substitute the coordinates of (3, 5) for x and y, obtaining the equation
(5 - (-3)) = a ( 3 - 1)^2, or{}(5 + 3) = a * 2^2, or{}8 = a * 4,
with the obvious solution
a = 2.
Thus the equation of the parabola is
y + 3 = 2 ( x - 1)^2.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
How do you know that it is y +3?
@&
The most direct answer is that
y + 3 = 2 ( x - 1)^2
is the same as
y = 2 ( x - 1)^2 - 3
so the points of this graph lie 3 units lower than the point of the graph of
y = 2 ( x - 1)^2,
which at its vertex has y = 0.
*@
------------------------------------------------
Self-critique Rating: 2
@&
This is also an application of the fact that the graph of
y - k = A ( x - h ) ^ 2
has its vertex at the point (h, k).
*@
*********************************************
Question: * Extra Problem (4.1.67 7th edition) (was 4.1.50). What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5?
Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3).
If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15.
If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30.
If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30.
If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75.
Does the value of a affect the location of the vertex?
In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1.
The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following:
For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12.
For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24.
For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24.
For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60.
So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60).
Self-critique (if necessary)
------------------------------------------------
Self-critique Rating: 3
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
&#Good responses. See my notes and let me know if you have questions. &#