course Mth 163 003. `query 3
.............................................
Given Solution: ** STUDENT RESPONSE: The graphs of all of the plots were parabolic with all of the vertices at (0,0). Graphs of the plots swtretched by factors .5, 2 and 3 all had the vertex 'down' with factor 3 graph as the narrowest, then factor 2 graph and finallyfactor .5 as the broadest. Factor -.3 was the broadest of them all and its vertex was up. INSTRUCTOR COMMENT: Good descriptions. You should also mention that the factor 3 graph contained the points (-1, 3) and (1, 3), while the factor 2 graph had points (-1, 2) and (1,2) while the factor .5 graph had (-1, .5) and (1, .5). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = x^2 + 2x + 1 x = - b / (2 a) x = -2 / (2(1) = -1 y = x^2 + 2x + 1 y = (-1)^2 + 2(-1) + 1 Y = 0 (-2,0) (-1,0) (0,0) y = x^2 + 3x + 1 x = - b / (2 a) x = -3 / (2) = -3/2 y = x^2 + 3x + 1 y = (-3/2)^2 + 3(-3/2) + 1 y = 2.25 -4.5 + 1 y = -1.25 (-2.5, -1.25) (-1.5, -1.25) (-0.5,-1.25) confidence rating: 3 OK ............ My answers are off, maybe? -1/4 is -.25 1.25 is 5/4
.............................................
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, 1.25) and (-.5, 1.25). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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: They move in a curve downward confidence rating: 3 OK OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow do the three fundamental points allow you to make a reasonably accurate sketch of the entire parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It shows the direction it opens and a vertex confidence rating: 3 OK OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER AND INSTRUCTOR COMMENT: They provide the initial sweep & symmetry around the vertex which defines the 'shape' and direction and allows you to extrapolate. INSTRUCTOR COMMENT: That's very nicely put. The vertex lies on a vertical line of symmetry and the points to right and left show you which way the parabola opens (upward or downward) and indicate the width of the parabola. ** ANOTHER VERY GOOD STUDENT STATEMENT: Vertex provides the anchor point and central line of symmetry, and the other two points give the direction and shape of the curve. STUDENT COMMENT: With my drawing skills, 3 points doesn't seem to be enough. However, the three fundamental points do provide the vertex, or in other terms, the line of symmetry or the parabola, as well as an additional point to each side of the vertex to determine the approximate shape of the curve, which would indicate whether the parabola opens upwards or downwards at the very least. INSTRUCTOR RESPONSE: You're not going to get a completely accurate picture, but you can see the direction in which the parabola opens and get a pretty good idea how wide or narrow it's going to be. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery Zeros of a quadratic function: What was it that determined whether a function had zeros or not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The 'sqrt function cannot be negative and get a zero point If 'sqrt is 0 then you will have 1 zero point If 'sqrt is 1 or greater you will have two zero points confidence rating: 3 OK OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER AND INSTRUCTOR COMMENT: Initially, I used the graph of my fundemental points as a visual aid as to whether zero crossings were present INSTRUCTOR COMMENT:There are two ways to answer this question. One is your answer--the vertex and two basic points clearly show you whether the function has zeros. The other way to answer is to note that if the discriminant b^2 - 4 a c, which appears inside the square root, is negative the function cannot have zeros (since you can't take the square root of a negative) and if the discriminant is positive you do have zeros (the quadratic formula will yield one zero if the discriminant is zero and will yield two zeros if the discriminant is positive). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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: Its uniform confidence rating: 3 OK OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat was the shape of the curve connecting the vertices? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Possible Parabola confidence rating: 3 OK OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. Query Add comments on any surprises or insights you experienced as a result of this assignment. I think some of it is coming together, I feel better about my work. "