#$&*
Mth 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Major Quiz Question 1
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I am looking over some of the test problems and I want to be very sure that my reasoning and process for determining the answers is correct. I will not ask for answers, rather I will show you the problem, and tell you how I would go about solving it, and allow you to clear up any discrepancies if this is alright.
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Problem: The quadratic function y = .029 t2 + -1.9 t + 53 models depth vs. clock time. Find the vertex of the graph and the graph points 1 unit to the right and the left of the vertex. Find the zeros of the function, and indicate all these points on a graph of the function.
Problem: What function do we get if we vertically stretch the basic y = x 2 parabola vertically by a factor of 2, then shift it -1.9 units vertically? Sketch a graph showing the effect of the vertical stretch on at least four selected points on the entire graph, then do the same for the vertical shift.
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For the first problem, the x value of the vertex of the graph is found by dividing -b/2a, the y value by substituting the x value into the quadratic function. I'm not sure how to find graph points 1 to the left and right except that the value of A in the quadratic function is our vertical shift so that tells us how far to go up, how do we determine where to go right or left? The zeros of the function are found by setting y = 0 and solving using the quadratic formula. Just to clarify, when setting y = 0 the minus/plus values of the function do NOT change? I would then sketch the graph and plot the points.
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The parabola is a stretched and shifted version of the basic parabola y = t^2, with its basic points (-1, 1), (0, 0) and (1, 1).
The first operation that would be applied to the y = t^2 parabola would be the vertical stretch by factor .029. This would give you the function y = .029 t^2, which multiplies the y value of all the points of the y = t^2 parabola by .029. In particular applying this to the basic points of y = t^2 we get the basic points (-1, .029), (0, 0) and (1, .029) of our vertically stretched parabola y = .029 t^2.
We would then shift the function in the x and the y directions so that the vertex moves from (0, 0) to the vertex obtained by letting x = -b / (2 a). The horizontal stretch will be -b / (2 a), and the vertical stretch would be the corresponding value of y.
These are shifts, not stretches, so the spacing of the three basic points will not change. The first basic point lies 1 unit to the left and .029 units vertically from the vertex, the second 1 unit to the right and .029 units vertically from the vertex.
The .029 is the value of a for this function. So to get the three basic points we put the vertex at the appropriate point, with the other two points lying 1 unit right and 1 unit left, both with a units in the vertical direction from the vertex.
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Your zeroes will be at
(-(-1.9) +- sqrt( (-1.9)^2 - 4 * .029 * 53) ) / (2 * .029).
Having figured out the three basic points and the zeroes, you can use a graphing calculator to verify your results.
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For the second problem, the vertical stretch factor is given as y = 2x^2, to shift it -1.9 units vertically we would say y = 2x^2-1.9. To show this i would need to take 4 selected points of y = x^2, then show how the points change for a vertical stretch of 2, and how they change for a vertical shift of -1.9. Should i have 2 graphs, (y=x^2 and y = 2x^2-1.9) or 3, (y = x^2, y = 2x^2, y = 2x^2 - 1.9)
@&
Right.
However you can base your graph on the three basic points (-1, 1), (0, 0) and (1, 1) of the basic y = x^2 parabola.
Vertical stretch 2 gives you the points (-1, 2), (0, 0) and (1, 2), which are on the graph of y = 2 x^2.
The vertical shift -1.9 then shifts these points to (-1, 2 - 1.9) = (-1, 0.1), (0, 0-1.9) = (0, -1.9) and (1, 2 - 1.9) = (1, 0.1).
If you plot these points and sketch a reasonable parabola you'll have a reasonable graph of the function y = 2 x^2 - 1.9.
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#$&*
Mth 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Major Quiz Question 2
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Problem: For the function y = f(t) = .026 t^2 + -2.5 t + 91, what are the values of the following: f( 26.87491) and f( t + .01 )? What equation would you solve to determine the value of t for which f(t) = 45.1366? (You need not actually evaluate the equation). What is the value of the function for clock time t = 40.31236?
Problem: Sketch a graph representing the linear function family y = m x + b for m = 1.05, with b varying over all negative real numbers.
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For the first problem, f( 26.87491) is inputted into the function and solved, f( t + .01 ) when inserted gives what looks like another quadratic function.
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f(t + .01) = .026 * (t + .01)^2 - 2.5 ( t + .01) + 91; you would then want to expand the square, apply the distributive law, and simplify. You will get another quadratic function, fairly similar to the first but with different values of b and c.
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THe question asks which of these equations would be best for solving f(t) = 45.1366. THe first function gives a value in the realm of 500, and the function derived from f(t+0.01) gives a lesser value of 31.
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The question doesn't ask which one of these would best apply to f(t) = 45.14. It asks you to give the equation you would solve to find the value of t, and you would solve the equation f(t) = 45.14. The equation would be .026 t^2 - 2.5 t + 91 = 45.14.
You solve this equation by subtracting 45.14 from both sides to get the standard form a t^2 + b t + c = 0, to which you then apply the quadratic formula.
One thing you might have been done to get a result around 500 is plug 45.14 in for t. That is not something you want to do. The question asked what value of t gives you 45.14 as the function value, not what the value of the function is if t = 45.14.
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How do we determine which is more appropriate?
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You weren't asked to determine which is more appropriate.
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To solve for clock time t = 40.31 we simply insert this into the function and solve for y. A linear graph of y = mx+b with a slope of 1.05 will show a line with points each having a slope of 1.05 to the next one. y = 1.05x will cross through the origin, adding any negative real number for B will change the y intercept by that value.
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Good.
So to graph the family you would mark off a number of points on the negative y axis, representing negative values of b, and sketch a set of parallel lines, each with slope 1.05, through these points.
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I'm uncertain about which equation to solve for a given value of T, stated above, and i'm uncertain what all b varying over all negative numbers entails.
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Good questions. Check my notes.
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#$&*
Mth 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Major Quiz Question 3
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Problem Number 3
Explain how the discriminant b2 - 4 a c tells us how many zeros a quadratic function has.
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The discriminant tells us that if the discriminant is less than zero, we have two imaginary numbers and no x intercepts, a discriminant value equal to zero will give 1 intercept, and a greater then zero value will give two x intercepts.
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This is the information I learned from working through various problems with quadratic formula, but there may exist a why here that I'm not answering. Just want to make sure that if i see this one on the test I covered all the bases.
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To give a full explanation you want to say a little more about 'why', as I'm sure you suspect.
The 'why' is fairly simple. The discriminant appears in the quadratic formula with the +-:
x = (-b +- sqrt(b^2 - 4 a c) ) / (2 a)
If the discriminant is negative the square root is not a real number, since the square of any real number is positive. So in this case there is no real solution and the graph doesn't cross the x axis.
If the discriminant is zero the square root is 0, and it doesn't matter whether you add or subtract 0 to -b; either way you get the same numerator, which is -b. So there is only one solution, and that is x = -b / (2 a). (Note that this puts the vertex on the x axis).
If the discriminant is positive then its square root is a positive number, and b + sqrt(b^2 - 4 a c) will be different than b - sqrt(b^2 - 4 a c). So in this case there are two solutions.
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