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# The powers of General Form - PowerPoint PPT Presentation

The powers of General Form. Probe. Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship. a) Find and correct the odd-one-out. b) Name the 5 forms. Probe.

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### The powers of General Form

Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship.

a) Find and correct the odd-one-out.

b) Name the 5 forms

Below are 5 different ways of representing a quadratic relationship. Four of them represent the SAME quadratic relationship.

a) Find and correct the odd-one-out.

b) Name the 5 forms

mapping rule

equation in general form

equation in transformational form

graphical (parabola)

table of values

General Form toTransformational Form

Let’s look at an example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:

1

We know that in transformational form the coefficient of ‘x’ is 1

STEP 1:Divide every term by a.

General Form toTransformational Form

Let’s look at an example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:

STEP 2:Move the non-x term over

We’re getting close, but the right-hand side of the equation needs to be a perfect square: (x – HT)2

So far we have: x2 + 6x

This looks like:

x2

x

x

x

x

x

x

x2

x

x

x

x

x

x

The equivalent to making x2 + 6x a perfect square is to arrange these tiles into a square shape...

x

x

x

x2

x

x

x

Oops…

x2

x

x

x

x

x

x

Try again…

Closer….

We just need tocomplete this square…

x2

x

x

x

x

x

We were missingnine 1×1 squares.

x

---------- x + 3 -----------

This looks promising:We have a square with length and width dimensions of (x + 3).

We just had to add 9, or 32.

---------- x + 3 -----------

x2

x

x

x

But how can we justify this “just add 9”?

Add it to both sides of the equation!

But why 32?

It’s half of the coefficient of x, squared.

x

x

x

General Form toTransformational Form

Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:

STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)

General Form toTransformational Form

Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:

STEP 4:Factor out the 1/a term on the left hand side

General Form toTransformational Form

Back to our example.We will take the quadratic function given in general form:y = 2x2 + 12x – 4and turn it into transformational form:

Therefore, the transformational form of the quadratic functiony = 2x2 + 12x – 4 is:

Now we know that its VS = 2, HT = –3, and VT = –22

From this we know its vertex, range, axis of symmetry, etc

vertex is (–3, –22)

range is{y≥–22, y R}

axis of symmetry isx = –3

1. Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.

2. Complete the square to find the range of the functiony = –x2 – 5x +1?

3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

Vertex: (2, –6)

Complete the square to find the vertex of the functiony = 2x2 – 8x + 2. Sketch its graph.

range is{y≤ 7.25, y R}

2. Complete the square to find the range of the functiony = –x2 – 5x +1?

3. Put the equation y = 0.5x2 – 3x – 1 into transformational form by completing the square.

STEP 1:Divide every term by a.

STEP 2:Move the non-x term over

STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)

STEP 4:Factor out the 1/a term on the left hand side

But completing the square is a lot of work….

Let’s find a shortcut to finding vertex ect. by completing the square just ONE more time, but with the general equation

STEP 1:Divide every term by a.

STEP 2:Move the non-x term over

STEP 3:Complete the square(take half the coefficient of x, square it and add it to both sides)

STEP 4:Factor out the 1/a term on the left hand side

We now see that ANY general quadratic equationy = ax2 + bx + c can be written as

This may look complicated, but is VERYhelpful…

from this we can see that the HT for any general quadratic is

For example: What is the axis of symmetry of the function y = –2x2 + 4x + 6?

To complete the square on this takes time. But…

We now see that ANY general quadratic equationy = ax2 + bx + c can be written as

This may look complicated, but is VERYhelpful…

from this we can see that the VT for any general quadratic is

For example: What is the vertex of the function y = –2x2 + 4x + 6?

To complete the square on this takes time. But…

vertex (1, 8)

We now see that ANY general quadratic equationy = ax2 + bx + c can be written as

This may look complicated, but is VERYhelpful…

from this we can see that the VS for any general quadratic is a.

Since the VS = –2 we can graph from the vertex (1, 8):

Over 1 down 2

Over 2 down 8

Over 3 down 18

For example: Graph the functiony = –2x2 + 4x + 6.

To complete the square on thistakes time. But…

Instead of memorizing the formula for the y-coordinate of the vertex (VT):

we can calculate it using the general form of the equation and thex-coordinate of the vertex (HT):

For example: What is the maximum value of the function y = –2x2 + 4x + 6?

Max value is y = 8.

For the following quadratic functions, find the vertex, sketch its parabola, give its axis of symmetry, give its range, and write in transformational form, all WITHOUT completing the square.

vertex (1, –2), VS = 1/4

axis of symmetry: x = 1range: {y≥ 2, y R}

General form shortcuts: PracticeSolutions

vertex (–4, –6), VS = 3

axis of symmetry: x = –4range: {y≥ –6, y R}

Solving for x given y using general form

Given a quadratic function in general form y = ax2 + bx + c and a value for y, it is not a straightforward task to find the corresponding x-value(s) because we cannot easily isolate x when it appears in 2 different forms: x and x2.

Subbing in the required y-value to a quadratic function, and bringing all the terms to one side of the equation yields a quadratic equation.

A quadratic equation in general form looks like this:ax2 + bx + c = 0 where a≠ 0.

Solving a quadratic equation for x is also calledfinding its roots or finding its zeros.

Solving for x given y using general form

NOTE: the c value of the quadratic equation might be different from the c value of the quadratic function. See the following example.

Example 1:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = –12.

Although the quadratic function had coefficients a = 1, b = –2, and c = –15,subbing in −12 for y gives the quadratic equation with coefficientsa = 1, b = –2, and c = –3

Solution:

−12 = x2 – 2x – 15

0 = x2 – 2x – 3

Solving for x given y using general form

NOTE: the c value of the quadratic equation is the SAME as the c value of the quadratic function when we are using y = 0. See the following example.

Example 2:Give the quadratic equation obtained from the functiony = x2 – 2x – 15 when y = 0.

Both the quadratic function and the resulting quadratic equation have coefficients a = 1, b = –2, and c = –15.

Solution:

0 = x2 – 2x – 15

While this is only true when we use y = 0, this is a very common case, since we are often interested in the x-intercepts of a quadratic function.

Solving for x-intercepts using general form

Example 3:Find the x-intercepts of the quadratic function y = x2 – 2x – 15

Solution:

0 = x2 – 2x – 15

…but now what?

Soon we will learn how to do this directly from general form…. Last class we learned how to solve this if the function had been given in transformational form…

So lets try that!

Solving for x-intercepts using general form

STEP 1: Write in transformational form by completing the square

STEP 2: Set f(x) = y = 0

STEP 3: Simplify left-hand side

STEP 4: Take the squarrootof both sides

STEP 5: Isolate x in both equations

Therefore the x-intercepts of the parabola are (5, 0) and (–3, 0), and the roots of the function are 5, and –3.

Solving for x-intercepts using general formPractice

Find the roots of the following quadratics:

Not worth it!

Wait for the

shortcut!

STEP 1: Write in transformational form by completing the square

STEP 2: Set f(x) = y = 0

STEP 3: Simplify left-hand side

STEP 4: Take the square-root of both sides

STEP 5: Isolate x in both equations

But this is a lot of work….

Let’s find a shortcut to finding x-intercepts by doing this just ONE more time, but with the general equation.

STEP 1: Write in transformational form by completing the square

STEP 2: Set f(x) = y = 0

STEP 3:Simplify left-hand side

STEP 4: Take the square-root of both sides

STEP 5: Isolate x in both equations

This is the shortcut, the Quadratic Root Formula!

Solving for x-intercepts using general formThe short-cut

Back to our example:Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15

Solution:

Instead of those 5 steps, let’s apply the Quadratic Root Formula.

Solving for x given y using general form

The quadratic root formula can be used to find x-values other than the x-intercepts

For the function f(x) = x2 – 2x – 15, find the values of x when y = −12:

Solution:

–12 = x2 – 2x – 15

0 = x2 – 2x – 3.

Now use the quadratic root formua witha = 1, b = –2 and c = –3

Solving for x-intercepts using general formPractice

Find the roots of the following quadratics:

Use the

root formula!

Find the x-intercepts of the following quadratic functions:

As slick as the quadratic root formula is for finding roots, sometimes factoring is even faster!

The zero property:

If (s)(t) = 0 then s = 0, or t = 0, or both.(This only works when the product is 0.)

So if we can get the quadratic function in the form:y = (x – r1)(x – r2), (called factored form)then when y = 0 we know that:

(x – r1) = 0, or (x – r2) = 0, or both.

In other words, x = r1 or x = r2 or both.

Ex. Find the x-intercepts of the graph of the functionf(x) = x2 – 2x – 15

Answer when the numbers are multiplied

We need to factor x2 – 2x – 15.We need 2 numbers whose product is –15 and whose sum is –2

(There is never more than one pair of numbers that work!)

-5

3

___ ×___ = –15

___ + ___ = – 2

-5

So when looking for roots, 0 = (x – 5)(x + 3) and x = 5 or x = –3

3

So the factored form off(x) = x2 – 2x – 15 isf(x) = (x – 5)(x + 3)

So the x-intercepts of the graph are (5, 0) and (–3, 0)

When factoring a quadratic equation where a = 1,find two numbers that multiply to give c and add to give b.

By the way:

(0) = x2 + 10x – 24

24 = x2 + 10x

24 + 25 = x2 + 10x + 25

49 = (x + 5)2

±7 = x + 5

x = –5 + 7 or x = –5 – 7

x = 2 or x = –12

Ex. Find the roots ofy = x2 + 10x – 24by factoring.

12

-2

___ x ___ = -24

___ + ____ = 10

-2

12

or

a) 0 = x2 + 2x +1

c) 0 = x2 + 2x – 24

b) 0 = x2 + 5x +4

d) 0 = x2 – 25

2. Find the x- and y- intercepts of the following quadratics.

a) f(x) = 2x2 +3x + 1

c) f(x) = x2 – 5x -14

b) f(x) = 3x2 – 7x + 2

d) y = 2(x – 4)2 – 32