Assignment 24

course Mth 158

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024. `query 24

College Algebra

08-06-2008

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00:17:58

3.4.14 (was 3.3.6). Concave down then concave up.

Does this graph represent a constant, linear, square, cube, square root, reciprocal, abs value or greatest integer function?

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RESPONSE -->

This was a cube function.

confidence assessment: 3

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00:18:21

** A linear function, represented most simply by y = x, has no curvature.

A quadratic function, represented most simply by y = x^2, has a parabolic graph, which is either concave up or concave down.

A cubic function, represented most simply by y = x^3, has a graph which changes concavity at the origin. This is the first power which can change concavity.

A square root function, represented most simply by y = sqrt(x), have a graph which consists of half of a parabola opening to the right or left. Its concavity does not change.

A reciprocal function, represented most simply by y = 1 / x, has a graph with a vertical asymptote at the origin and horizontal asymptotes at the x axis, and opposite concavity on either side of the asymptote.

An absolute value function, represented most simply by y = | x |, has a graph which forms a 'V' shape.

The greatest integer function [[ x ]] is a 'step' function whose graph is made up of horizontal 'steps'.

The concavity of the reciprocal function changes, as does the concavity of the cubic function. The cubic function is the only one that matches the graph, which lacks the vertical and horizontal asymptotes. **

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RESPONSE -->

ok

self critique assessment: 3

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00:19:17

3.4.20 (was 3.3.12).

Does your sketch of f(x) = sqrt(x) increase or decrease, and does it do so at an increasing or decreasing rate?

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RESPONSE -->

It increases at a decreasing rate.

confidence assessment: 2

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00:19:29

** y = sqrt(x) takes y values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4.

sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2.

The change in y from the x = 0 to the x = 2 point is about 1.414; the change from the x = 2 to the x = 4 point is about .586. The y values therefore increase, but by less with each 'jump' in x.

So the graph contains points (0,0), (2, 1.414) and (4, 2) and is increasing at a decreasing rate. **

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RESPONSE -->

ok

self critique assessment: 3

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00:20:10

What three points did you label on your graph?

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RESPONSE -->

(0,0) (1,1) (4,2)

self critique assessment: 3

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00:39:49

3.4.24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0.

Given the intercepts, domain and range of the function

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RESPONSE -->

y-intcepts are (-2.5, 0) and about (-1.5, 0)

x-intercepts are (0, 5) and (0, -3)

Domain [all real numbers}

Range {yly>-3, all real numbers}

confidence assessment: 1

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00:40:11

** From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at

2x + 5 = 0 or x = -5/2.

Since this x value is in the interval from -3 to 0 it is part of the graph.

At x = -3 we have y = 2 * (-3) + 5 = -1.

So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5).

The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5).

For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin.

For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). **

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RESPONSE -->

ok

self critique assessment: 3

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&#Very good responses. Let me know if you have questions. &#