Assignment 3 Query

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course Mth 163

003. `query 3

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Question: `qquery graph of y = x^2 stretched vertically by different factors. When you constructed a graph of y = x^2 and stretched it, describe the graphs you obtained when you stretched it vertically by a factor of 2, by a factor of 3, by a factor of .5 and by a factor of -.3.

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Your solution:

All the graphs were parabolic and vertex of (0,0)

confidence rating #$&*:

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Given Solution:

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Question: `qquery prob 3: Location of the vertex and fundamental points. Determine the location of the vertex and fundamental points of each of the following quadratic functions: y = x^2 + 2x + 1 and y = x^2 + 3x + 1

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Your solution:

The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a). one fundamental point is the vertex (-3/2, -5/4).

confidence rating #$&*:

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Given Solution:

** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a).

For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0).

The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-2,1) and (0,1).

For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4).

The other two fundamental points will lie 1 unit to left and 1 unit to the right of the vertex and will be vertically displaced by a = 1 units from the vertex, so these points will be (-5/2, -1/4) and (-1/2,-1/4); or in decimal form at (-2.5, -.25) and (-.5, -.25). **

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Question: `qhow did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?

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Your solution:

The vertices move down and to the left.

confidence rating #$&*:

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Given Solution:

** The vertices move downward and to the left, but not along a straight line.

In fact the vertices lie along a different parabola of their own.

You wouldn't have been expected know at this point that the vertices all lie along a parabola of their own, of course, though students occasionally conjecture that it is so. However that statement should make sense in terms of your picture, and it's a very interesting and unexpected connection.**

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Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?

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Your solution:

They provide the structure needed to the vortex

confidence rating #$&*:

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Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not?

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Your solution:

If the discriminant is negative the function cant have 0

confidence rating #$&*:

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Question: `qquery #4. Questions about vertex between zeros and the shape of the curve connecting vertices:

What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?

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Your solution:

Graphs of functions are symmetric throught the vertex. If there is a point on the right of the vertex a point on the opposite side must be the same.

confidence rating #$&*:

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Given Solution:

** The graph of each function is completely symmetric about a vertical line through the vertex. Thus if one of the zeros lies to the right of the vertex, the other must lie an equal distance to the left of the vertex. **

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Question: `qWhat was the shape of the curve connecting the vertices?

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Your solution:

round

confidence rating #$&*:

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Given Solution:

** In fact the vertices of y = x^2 + b x + 1 for various values of b (e.g., b = 2 and b = 3 for the parabolas of the preceding problem) themselves lie on a parabola. We're a ways from actually proving that, but it's an interesting idea.

Don't actually do it but if you took any three of these vertices and fit them to the form y = a t^2 + b t + c you would get an equation for y vs. t, and all the other vertices would satisfy the equation. **

The figure below shows the graphs of the above functions, which have the form y = x^2 + b x + 1 for b = 2 and 3 (i.e., when b = 2 the form y = x^2 + b x + 1 becomes y = x^2 + 2 x + 1; when b = 3 the form gives us x^2 + 3 x + 1), as well as all the functions we get for b = -5, -4, -3, ..., 3, 4, 5.. You should be able to identify the parabolas you graphed among these.

The figure below depicts the same graphs, and includes as well the parabola that passes through the vertices of all the other parabolas. The equation of that new parabola was obtained by using the vertices (0, 1), (-1, 0) and (-1.5, -1.25) of the b = 0, b = 2 and b = 3 parabolas. Using those vertices as our three selected points, we could easily write down then (tediously) solve the equations to get the parabolic model that fits the points.

Self-critique (if necessary):

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Self-critique rating:

Assignment 3 Query

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course Mth 163

003. `query 3

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Your solution:

All the graphs were parabolic and vertex of (0,0)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Your solution:

The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a). one fundamental point is the vertex (-3/2, -5/4).

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The vertex of y = a x^2 = b x + c occurs at x = - b / (2 a).

For y = x^2 + 2 x + 1 we have vertex at x = -2 / (2 * 1) = -1. At x = -1 we have y = (-1)^2 + 2 * -1 + 1 = 0. So one fundamental point is the vertex (-1,0).

For y = x^2 + 3 x + 1 we have vertex at x = -3 / (2 * 1) = -3/2. At x = -3/2 we have y = (-3/2)^2 + 2 * (-3/2) + 1 = 1/4. So one fundamental point is the vertex (-3/2, -5/4).

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Question: `qhow did the location of the vertex change as you moved from the graph to graph, for the first four graphs given in the problem?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The vertices move down and to the left.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

** The vertices move downward and to the left, but not along a straight line.

In fact the vertices lie along a different parabola of their own.

*********************************************

Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

They provide the structure needed to the vortex

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the discriminant is negative the function cant have 0

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Question: `qquery #4. Questions about vertex between zeros and the shape of the curve connecting vertices:

What indication do your graphs give you that the vertex of a parabola lies on a line halfway between its two zeros, provided these zeros exist?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Graphs of functions are symmetric throught the vertex. If there is a point on the right of the vertex a point on the opposite side must be the same.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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Question: `qWhat was the shape of the curve connecting the vertices?

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Your solution:

round

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

Self-critique (if necessary):

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Self-critique rating:

#*&!

@&
You appear to be deleting some of the text of the original file.

I rely on codes within these documents to quickly separate your answers from the rest of the text, and cannot do so without those codes.

So I'm going to have to ask you to make another copy of the document and insert your solutions into that copy. You can insert information into a document, but don't delete anything that was there in the first place.

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