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.

 

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:

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

Solving quadratics with inequalities

Q:  Solve  (a/2)2 – a > 0

A:  Follow the steps below

Solving Algebraically:

Step 1:  Simplify

We can simplify the equation by squaring the fraction to get:

a2/4 – a > 0

OR in different form:

(1/4) a2 – a > 0

Now, to get rid of the fraction and clean things up, I am going to multiply everything by 4 (since the fraction is 1/4 — you don’t have to do this, just a “cleaning up” step”):

4*(1/4) a2 – 4*a > 4*0

a2 – 4a > 0

Normally, when you solve inequalities, you isolate the variable by moving things around to the left / right side.  When it is a quadratic, you don’t want to do that.  You want all of the numbers and variables on one side and zero on the other side.  We have that, so we are good to go.

Step 2:  Factor

Now, factor the left side:

a2 – 4a > 0

a (a – 4) > 0

Step 3:  Identify the zeroes

As we are used to doing with quadratics, we need to find what values you plug in to make “zeroes”

So, take each factor and set it equal to zero like so:

a = 0 and a – 4 = 0

Solve to get:

a = 0 and a = 4

The zeroes are:  0 and 4.

So, this parabola (quadratic) crosses the x-axis at 0 and 4.  Now, we want to find where the quadratic is greater than 0.

With analysis, we know the quadratic is concave up in shape (a smiley face).  On your paper, draw an x-y graph with a parabola that crosses the x-axis and 0 and 4 and is concave up.  You should get:

 

So, where is the parabola > 0 (above the x-axis?)

When a < 0 and when a > 4

 

Graphing Quadratics

Q:  Sketch graphs of these three quadratics relations on the same set of axes

a) y=-3x2

b) y=1/4 x2
A:  There are many ways to sketch graphs (quadratics).  The most basic way is to:  (1) make a table of values, (2) plot the points, (3) connect the dots.

So, let’s start with the first equation (a) y=-3x2 and make a table of values:

STEP 1

Start by picking values for x.  I will pick -2, -1, 0, 1, 2 for x:

x  | -2 | -1 | 0 | 1 | 2

y  |     |     |      |    |

Now we need to calculate the y values.

When x = -2, y=-3(-2)2 = -3(4) = -12

When x = -1, y=-3(-1)2 = -3(1) = -3

When x = 0, y=-3(0)2 = -3(0) = 0

When x = 1, y=-3(1)2 = -3(1) = -3

When x = 2, y=-3(2)2 = -3(4) = -12

So, the table is complete:

x  | -2   |  -1  | 0   | 1   | 2

y  | -12 |  -3 |  0  | -3 |  -12

STEP 2:

Plot the points: (-2, -12)  (-1, -3)  (0, 0)  (1, -3)  (2, -12)  on the axes

STEP 3

Connect the dots to make a parabola.  You should have a picture like so:

Now we need to do the same thing for equation (b) y=1/4 x2

STEP 1

Start by picking values for x.  I will pick -2, -1, 0, 1, 2 for x:

x  | -2 | -1 | 0 | 1 | 2

y  |     |     |      |    |

Now we need to calculate the y values.

When x = -2, y=1/4 (-2)2 = 1/4 (4) = 1

When x = -1, y=1/4 (-1)2 = 1/4 (1) = 1/4

When x = 0, y=1/4 (0)2 = 1/4 (0) = 0

When x = 1, y=1/4 (1)2 = 1/4 (1) = 1/4

When x = 2, y=1/4 (2)2 = 1/4 (4) = 1

So, the table is complete:

x  | -2   |  -1    | 0   |   1   | 2

y  |  1    |  1/4 |  0  | 1/4 |  1

STEP 2:

Plot the points: (-2, 1)  (-1, 1/4)  (0, 0)  (1, 1/4)  (2, 1)  on the axes

STEP 3

Connect the dots to make a parabola.  You should have a picture like so:

SO…. If we wanted to put these two graphs on the same axes, it would look something like (depending on your scale on the y-axis):

Graphing a Quadratic Function

Q:  Graph the Quadratic Function:  f(x) = 2x2 – 3x + 1

Answer:

There are two main methods to do this.  I will do both methods (depending on what you have learned in class, you will want to pick either method 1 or method 2 that I show).

Method 1:  Plotting Points

Because this is a quadratic equation, we know the graph will be in the shape of a parabola.  So, I am going to pick a few values for x, then find the y values.  Then, I will plot the points on the graph to make a parabola.  You can pick any values for x when you do this method.

Let’s pick x = 0.  So, plug x into the equation to find y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

Point 1:  When x = 0, y = 1:  (0, 1)


Let’s pick x = 1 now:

2(1)2 – 3(1) + 1= 2(1) – 3 + 1 = 2 – 3 + 1 = 0

Point 2:  When x = 1, y = 0:  (1, 0)


Let’s pick x = -1 now:

2(-1)2 – 3(-1) + 1= 2(1) + 3 + 1 = 2 + 3 + 1 = 6

Point 3:  When x = -1, y = 6:  (-1, 6)

Now, plot the points (0, 1) and (1, 0) and (-1, 6) on a graph.  Connect the dots to make a parabola!

Keep in mind:  You can get more than 3 points for more accuracy.

Method 2:  Finding the y-intercept and the vertex:

The y-intercept of any type of graph occurs when you plug in 0 for x.  So, plug in 0 for x into the equation and solve for y:

2(0)2 – 3(0) + 1 = 0 – 0 + 1 = 1

So, the y-intercept is the point (0, 1).

Now, there is a formula to find the vertex of a quadratic equation.  It goes as follows:

Step 1 to find the vertex:  x = -b / 2a (this will give us the x coordinate of the vertex.

Recall, in our equation, a = 2, b = -3 and c = 1

So, the x part of the vertex is x = -(-3) / 2(2) = 3/4

Now, to get the y part, you need to plug the x part into the equation and solve:

2(3/4)2 – 3(3/4) + 1 = -1/8


The vertex is the point (3/4 , -1/8)

So, on your graph, plot the vertex and the y-intercept.  Draw a parabola through the y-intercept until you hit the vertex.  You know the vertex is either the high or the low point.  Then, “bounce off” the vertex and continue drawing the other half of the parabola.

 

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.

This is factorable, so I am going to solve it by factoring.  Keep in mind, if you have never factored something this complicated, you will need to plug it into the quadratic formula (no big deal, just use a = 4, b = -7, c = 3 and plug it in to the big ol’ formula you have).

To factor… set up your parenthesis:

(            )*(              )

The first two terms need to multiply to give 4n2 and the back two terms need to multiply to give + 3.  To multiply and get +3 we either need 2 positive numbers or two negative numbers.  Specifically, we need -1 * -3 or 1*3.

Since we have a -7n in the middle, the back two terms both need to be negative.  Therefore, the two back terms must be -1 and -3:

( ___ – 1 )*( ___ – 3)

Now we need to figure out the front terms.  They must multiply to give 4n2 .  This leaves us with either 4n*1 or 1*4n or 2n*2n.

So you can try each combination in place, distribute it out by hand or visually to see which one will “work”.  Factoring does involve a decent amount of guessing and checking if you want to be good at it.

Through a little guessing and checking, I get:

4n2 – 7n + 3 = (n – 1)*(4n – 3) = 0

Which means either:

n – 1 = 0  or  4n – 3 = 0

Solve for n in each case to get:

n = 1 or n = 3/4

You will get the same answers if you had used the quadratic equation to solve (which may be easier if you are not comfortable with factoring more complicated expressions)