Compilation of Notes for Midterm I

 Completing the Square Step by step process of completing the square:  Example

2x² - 8x + 2

1. Factor the leading coefficient:  2(x² - 4x + 1)
2. Calculate -b/2:  -(-4)/2 = 2
3. Square the solution above:  2²  = 4
4. Add and subtract answer from part three inside parentheses: 2(x² - 4x + 4 - 4 + 1) 
5. Regroup:  2[(x² - 4x + 4 ) + 3]
6. Factor the inner parentheses using part two as a hint:  2[(x - 2)² +3]
7. Multiply out the outer constant:    2(x - 2)² + 6
8. Breath a sigh of relief.

 Applications

Some pointers on solving word problems.

1)  Read the question and ask yourself what the problem is asking.

2)  Label variables:  Write "Let x = ..."  where x should be a numerical variable that will give the solution.  Then label any other variables if needed

3)  If possible draw the picture.  Recall the Pythagorean Theorem:  a² + b² = c²  where and b are the lengths of the legs of the right triangle and c is the length of the hypoteneuse.

4)  Reread the problem phrase by phrase, inserting the variables when they are spoken in the phrase.  (Recal that is ,will be, was, are, equate to =.

5)  Using the phrases come up with appropriate equations.

6) Solve the equations.

7)  Reread the problem and answer it making sure that the answer makes sense.

Radical Equations

1)  Isolate the radical so that the radical is alone on the left side of the equation with everything else on the other side of the equation.

2)  Square both sides of the equation

3)  Bring everything back to the left side of the equation so that the right sid is a 0.

4)  Solve the resulting equation by factoring, the root method, the quadratic equation, of some other method.

5)  Plug your answer back into the original equation making sure that your result is an actual solution and not an extraneous one.

Hidden Quadratics:

Often we encounter an equation that is a quadratic in disguise.  For example the equation

x⁴  + 3x²  -  4 = 0

is a fourth degree equation.  Notice that the exponent of the leading term is twice the exponent of the second term.  Because of this we can use what is called a substitution:

Let u = x² then x⁴  = u²   

Then the equation in terms of u becomes:

u²   + 3u - 4 = 0

which is a quadratic.  Hence, we can factor:

(u - 1)(u + 4) = 0

So that u = 1 or u = -4

Since u = x²  we have that

x²  = 1 or x²  = -4

The first has solution x = 1 or x = -1  and the second has no solution.

 Quadratic Inequalities

We will solve using the following:

1)  Put everything on the lift hand side so that we have for example Quad > 0

2)  Factor and set equal to zero.

3)  Solve and place answers on a number line.  This will cut the number line into two or three regions.

4)  Pick a test value for each region and plug that test value into each of the factors.  Put plusses or minuses over the region depending on whether the test values test positive or negative.

5)  If the region has two plusses or two minuses the the region is positive

If the region has one of each then the region is negative.

6) If the inequality is "<" then include the negative regions

If the inequality is ">" then include the positive regions

7) If the inequality is a less (greater) than or equal to then include the endpoints with solid dot and the interval []

If the inequality is a less (greater) than then do not include the endpoints by showing an open dot and the interval ()

 Rational Inequalities

To solve rational inequalities, we can use the same technique that we used for  quadratic inequalities with the following adjustments:

A)  After putting everything on the left hand side we put the left hand side over a comon denominator.

B)  Instead of factoring and solving to find the cut points, we just set the numerator and denominator each equal to zero.

C)  The cut point that is determined from that denominator will never be included, thus will be bordered by a ( and shown as an open dot.

 Circles

Recall that a circle with radius r and center (h,k) is defined by the set of point of distance r from the point (h,k).  If (x,y) is on the circle then the distance from (h,k) to (x,y) is r.  The distance formula tells us that

(x - h)²  + (y - k)² = r²

Graphing Parabolas

Example:  sketch the graph of the parabola y =  x² - 4x -5

Step 1:  Find the x-coordinate of the vertex-  -b/2a

x = 4/2 = 2

Step 2:  Find the y-coordinate of the vertex by plugging the x coordinate into the equation.

y = 2² -4(2) - 5 = -9

Therefore the vertex has coordinates (2,-9)

Step 3:  Find the y-intercept by plugging in 0 for x

y = 0² - 4(0) - 5 = -5

Step 4:  Find the x-intercepts by setting y = 0 and factoring or quadratic formula.

0 = x² - 4x - 5

(x - 5)(x + 1) = 0

x = 5 or x = -1

Step 5:  If necessary, plug in more values of x to find a few additional points.

(Here it is not needed since we already have four points:  (2,-9), (0,-5), (5,0), (-1,0))

If steps 1 through 4 produce fewer than 3 points, it is recommended to plot a few additional points.

Step 6:  Graph it!

Key to the Practice Midterm

Problem 1

(-3,1]u(4,infinity)

Problem 2

sqrt5, -sqrt5

Problem 3

(-infinity,3)

Problem 4

A.  0  

B. The first graph is a funciton since it passes the VLT, but the second is not, since it fails the VLT

Problem 5

A.  -4x² -4x + 3

B.  4 - 2x - h

C.  -5

D.  7

E.  (7-14x)/(7 + 4x - x²)

Problem 6

The dimensions of hte box are 10x10x5

Problem 7

Vertex = (-1, -14), y-int = (0,15), x-ints = (-5,0) and (3,0)

The graph will be shown in class