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Quadratics: Standard Form to Vertex Form

Q:  Write the equation of y = -x2 + 2x + 2 in the form of y = a(x-h)2 + k

A:  To do this, we are going to use a strategy called “completing the square” — a fairly complicated algebra 2 concept.

We are currently in the form y = ax2+bx+c and we want to get to y = a(x-h)2 + k

First, let’s establish a, b, and c in our equation: y = -x2 + 2x + 2

a = -1, b = 2, c = 2

I will walk you through the steps on how to do the problem.  At the end, I will provide some explanation behind the concept and the “why”.

Step 1)  Divide the entire equation by “a”

So, divide everything by -1:

y = -x2 + 2x + 2

y/ -1 = (-x2 + 2x + 2) / -1

And simplify to get:

-y = x2 – 2x – 2

Step 2)  “Complete the Square”

This step involves finding what number to add (or subtract) into the equation that will make the x terms factor nicely.  To find this number, we follow a simple pattern:

Take 1/2 of the coefficient in front of the x term and then square it.

The coefficient in front of x is -2.

1/2 * -2 = -1

Square -1 to get +1.

The number we need to “Complete the Square” is +1.

Add this number to both sides of the equation (remember to do it to BOTH SIDES to keep the equation balanced).  Also, add in by the x’s just for ease, like so:

-y + 1 = x2 – 2x + 1 – 2

So, our modified equation is:

-y + 1 = x2 – 2x + 1 – 2

Step 3)  Factor the x terms:

We are now going to factor the part of the equation I’ve highlighted.  The part we factor involves the x’s and the “complete the square” number that we added to the equation:

-y + 1 = x2 – 2x + 1 – 2

Factoring x2 – 2x + 1 gives (x – 1)(x – 1) OR (x – 1)2

So, we now have:

-y + 1 = (x – 1)2 – 2

Step 4) Isolate y

We have finished the hardest part (completing the square)!  Now, we just solve for y and we are done:

-y + 1 = (x – 1)2 – 2

Subtract 1 from both sides:

-y = (x – 1)2 – 3

Multiply everything by -1:

y = -(x – 1)2 + 3

And there it is!  We have taken the original equation: y = -x2 + 2x + 2 and re-written it in a different form: y = -(x – 1)2 + 3


Some logic behind the process:

Completing the square is the process of figuring out “what number” is needed to add (or subtract) in the equation so that it will factor easily into something like (x-h)2.  Once we determine the number needed, we add it to both sides of the equation to maintain balance.  Remember, we aren’t altering the actual equation, we are just changing its appearance.  Once we found the correct number, the equation will factor the way we need it to.

 

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Solving logarithmic equations

Q:  Solve
log16x + log4x + log2x = 7

A:  It is easier (though not always necessary) to have all of the logs in the same base to proceed.  Since that is not the case, we need to use the change of base formula to put all of the logs into the same base.

Change of base formula says:
logab = logcb / logcwhere c can be any base of your choosing

What base should we go to?  Since 16, 4 and 2 can all be formed by powers of 2, let’s go to base 2:

log16x + log4x + log2x = 7

log2x / log216 + log2x / log24 + log2x / log22 = 7

Now, we can simplify:

log2x / 4 + log2x / 2 + log2x / 1 = 7

1/4 log2x + 1/2 log2x + log2x = 7

7/4 log2x = 7

log2x = 7 *(4/7)

log2x = 4

(remember, this reads:  the power you put on 2 to get x is 4)

x = 24

x = 16

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Solving Exponential Equations by Factoring

Q:  Solve for x:
22x + 2x+2 – 12 = 0

A:  The first thing we need to notice (from practice and experience) is that we can re-write this like so using our knowledge / rules of exponents:

(2x)2 + 22*2x – 12 = 0
Now simplify a little:

(2x)2 + 4*2x – 12 = 0

So, look at this in a new light.  What if we substitute each 2x with y?

(2x)2 + 4*2x – 12 = 0   turns into   y2 + 4*y – 12 = 0

(this isn’t necessary, but is helpful for visualization)

We see this is in the form of a quadratic and can be factored:

(y + 6)(y – 2) = 0

So, y = -6 or y = 2

Remember, y was a substitution for 2x. So, we really have:
2x = -6   or   2x = 2

Solve each equation separately.  Let’s start with:

Continue reading Solving Exponential Equations by Factoring

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Solve the system of equations (conics)

Q:  Solve the system of equations (both of which are conic sections)
1:   x2 + y2 – 20x + 8y + 7 = 0
2:  9x2 + y2 + 4x + 8y + 7 = 0

A:  I am going to solve this by using the elimination method (since I see that I can cancel out all of the y’s)

I am going to multiply equation 1 by (-1) and then add it to equation 2:

2:  9x2 + y2 + 4x + 8y + 7 = 0

1: -x2 – y2 + 20x – 8y – 7 = 0   [after it has been multiplied by -1]

Now, add them together to get equation 3:

3:  8x2 +24x = 0

Continue reading Solve the system of equations (conics)

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Negative Exponent Examples and Basics

Q:  What is a negative exponent?  What does a negative exponent do?

A:  A negative exponent is just notation that can be used to represent a reciprocal.  Let me show you some examples:

x – 1 = 1/x1 or 1/x

x-2 = 1/x2

4 – 1 = 1/41 or 1/4

4 – 2 = 1/42 = 1/16

Those are the basics.  Let’s look at a few more complicated examples of how negative exponents affect an expression:

3x – 4 = 3/x4

4x6y – 3 = 4x6/y3

OR… If the negative exponent is affecting something in a denominator, it will move it to the numerator like so:

1/x – 5 = 1*x5 = x5

6/y – 7 = 6y7

These are just a few basic examples to show you how negative exponents work. Of course things can get more difficult.

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Solving Quadratics with Inequalities

Q:  Solve for x:  (x + 3)(x – 4) < 0

A:

Step 1:  Find the zeroes

Since the quadratic is already factored, this isn’t too tricky.  If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).

OK, so what are the zeroes?

x + 3 = 0 or x – 4 = 0

The zeroes are x = -3 or x = 4

So, draw a number line and plot the zeroes on the number line:

numberline

Now, you have to test each “interval” that is separated by the “zeroes”.  There are three intervals to test.

Interval 1:  The numbers to the left of -3  –> in interval notation this is (-infinity, -3)

Interval 2:  The numbers between -3 and 4  –> in interval notation this is (-3, 4)

Interval 3:  The numbers to the right of 4  –> in interval notation this is (4, infinity)

Step 2:  Test each interval

Pick any number on interval 1 and test it into the original inequality.  I’ll pick -5:

(x + 3)(x – 4) < 0

(-5 + 3)(-5 – 4) < 0

(-2)(-9) < 0

18 < 0      <—  this is false, so numbers on this interval [interval 1] are not part of the solution.

Pick any number on interval 2 and test it into the original inequality.  I’ll pick 0:

(x + 3)(x – 4) < 0

(0 + 3)(0 – 4) < 0

(3)(-4) < 0

-12 < 0      <—  this is true, so numbers on this interval [interval 2] are a part of the solution.

Pick any number on interval 3 and test it into the original inequality.  I’ll pick 5:

(x + 3)(x – 4) < 0

(5 + 3)(5 – 4) < 0

(8)(1) < 0

8 < 0      <—  this is false, so numbers on this interval [interval 3] are not part of the solution.

SO:  The only interval that “worked” was interval 2.  Therefore, the solution is all number between -3 and 4.

In interval notation, we write that like: (-3, 4)

In inequality notation, we write that like: -3 < x < 4

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Combining Fraction (involving variables)

Q:  Condense:  3/2x + x/(2x-6)

Answer:

In order to add fractions, we need to have common denominators.  Remember, in order to get common denominators, we need to multiply by something.. Just because one denominator is 2x – 6 and the other is 2x, we cannot just subtract 6 to make them match.

So, the fraction on the left needs a (2x-6) and the fraction on the right needs a (2x)…  So, you have to multiply the fraction on the left by (2x-6) and the fraction on the right needs to be multiplied by (2x).  Remember:  multiply to the top and the bottom like so:

3/2x + x/(2x-6)

3(2x-6)/2x(2x-6) + x(2x)/(2x-6)(2x)

Now, simplify each numerator, leave the denominators alone:

Continue reading Combining Fraction (involving variables)

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Solving a quadratic

Q:  Solve for n:

4n2 + 3 = 7n
 

Answer:  Since there is an n-squared term, this is a quadratic equation.  In order to solve this, we need to set the whole equation equal to 0 first (so, let’s subtract the 7n over to the left side of the equation):

We get:

4n2 – 7n + 3 = 0  [notice that I put the n’s in order of n-squared, n, and then the constant 3]

Now, there are a few methods you may have learned to can help you solve this:  1)  Factoring or 2) Quadratic Formula or 3) Completing the Square.

Continue reading Solving a quadratic

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Venn Diagram Word Problem

Q:  Superburger sells hamburgers with the choice of ketchup, mustard and relish. One day they sold 256 hamburgers; 140 had mustard, 140 had ketchup, 84 had ketchup and relish, 62 had mustard but no relish, 68 had ketchup and mustard, 38 had all three condiments and 20 had none.

(a) The number sold with relish only is?

(b)The number sold with no relish is?

Continue reading Venn Diagram Word Problem

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End Behavior

1.  Find the end behavior of the following functions:

(a)  Consider:  y = (x²-2x)/(-x³-5x²+4)
as x goes to ∞, y goes to ?
as x goes to – ∞, y goes ?

(b) Consider y = (2x+1)/(-4x+1)
as x goes to ∞, y goes ?
as x goes to – ∞, y goes ?

Continue reading End Behavior