Quiz

course Mth 163

The question that I got was Give the quadratic formula and the answer is x=-b+'sqrt(b^2-4ac)/2a and x=-b-'sqrt(b^2-4ac)/2a.

The full statement of the quadratic formula contains words as well as symbols; you also need to be careful of signs of grouping. The way you've written it the b term is not divided by the denominator, and the denominator is 2, rather than 2a. However it's clear that you know the formula; your main omission is the words that go with it.

Review this in the worksheets, where the wording is stated very explicitly with instructions to remember the exact wording.

This question often comes up on tests, and if you know the wording it's easy credit. If you don't, then you lose important points.

004.

Precalculus I

01-29-2009

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assignment #004

004.

Precalculus I

01-29-2009

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21:29:25

`q001. Note that this assignment has 4 questions

If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.

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RESPONSE -->

f(3)=13, f(7)=53 and f(-5)=29...The graph would be a parabola with the function of a=1 b=0 and c=4 the vertex would be at 0

confidence assessment: 3

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21:30:14

f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain

f(3) = 3^2 + 4 = 9 + 4 = 13.

Similarly we have

f(7) = 7^2 + 4 = 49 + 4 = 53 and

f(-5) = (-5)^2 + 9 = 25 + 4 = 29.

Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4.

The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.

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RESPONSE -->

self critique assessment: 3

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21:36:26

`q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.

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RESPONSE -->

f(a)=a^2+4...f(x+2)=(x+2)^2+4...f(x+h)=(x+h)^2+4...f(x+h)-f(x)=(x+h)^2+4-(x^2)+4 or simplified h^2...and [f(x+h)-f(x)]/h=(x+h)^2+4-x^2+4/h or simplified h

confidence assessment: 3

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21:38:04

If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a:

f(a) = a^2 + 4.

Similarly to find f(x+2) we replace x with x + 2:

f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8.

To find f(x+h) we replace x with x + h to obtain

f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4.

To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h):

f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2.

To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that

[ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.

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RESPONSE -->

I dont understand where the 2x came from in the last two because I thought that they would cancel out

self critique assessment: 3

2 x h appears only once in the numerator, in the expression from f(x+h). It does not appear in the expression for f(x). There is no reason to expect that it cancels out. When the numerator is divided by h, it's this term that leaves 2x, and this term is all that's left if you let h approach zero.

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21:41:32

`q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.

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RESPONSE -->

f(x1)=5x1+7...f(x2)=5x2+7 and [f(x2)-f(x1)]/(x2-x1)=(5x2+7)-(5x1+7)/(5x2+7-5x1+7)=1

confidence assessment: 3

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21:43:01

Replacing x by the specified quantities we obtain the following:

f(x1) = 5 * x1 + 7,

f(x2) = 5 * x2 + 7,

[ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1).

We can factor 5 out of the numerator to obtain

5 ( x2 - x1 ) / ( x2 - x1 ) = 5.

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RESPONSE -->

I dont understand why the denominator didnt have the 5 in it??

The denominator in the expression [ f(x2) - f(x1) ] / ( x2 - x1 ) is ( x2 - x1 ).

The 5 is part of the expression for f(x). It's not part of the expression (x2 - x1).

self critique assessment: 3

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21:45:36

`q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?

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RESPONSE -->

3=5x+7 and solve for x and you get x=-2

confidence assessment: 3

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21:45:44

If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation

5x + 7 = -3.

We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.

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RESPONSE -->

self critique assessment: 3

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Quiz

2 x h appears only once in the numerator, in the expression from f(x+h). It does not appear in the expression for f(x). There is no reason to expect that it cancels out. When the numerator is divided by h, it's this term that leaves 2x, and this term is all that's left if you let h approach zero.

The denominator in the expression [ f(x2) - f(x1) ] / ( x2 - x1 ) is ( x2 - x1 ).

The 5 is part of the expression for f(x). It's not part of the expression (x2 - x1).

Good work and good questions. See my responses.

&#Let me know if you have questions. &#

Quiz

2 x h appears only once in the numerator, in the expression from f(x+h). It does not appear in the expression for f(x). There is no reason to expect that it cancels out. When the numerator is divided by h, it's this term that leaves 2x, and this term is all that's left if you let h approach zero.

The denominator in the expression [ f(x2) - f(x1) ] / ( x2 - x1 ) is ( x2 - x1 ). The 5 is part of the expression for f(x). It's not part of the expression (x2 - x1).

Good work and good questions. See my responses.

&#Let me know if you have questions. &#

The denominator in the expression [ f(x2) - f(x1) ] / ( x2 - x1 ) is ( x2 - x1 ).

The 5 is part of the expression for f(x). It's not part of the expression (x2 - x1).