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Math - Create Quadratic Equation

by KIM
(MIAMI ,FL )










































Create your own quadratic equation:

   1. that has an axis of symmetry of x = 1

   2. whose graph opens down.

Then find:

   a. the vertex

   b. domain

   c. range

   d. x-intercepts.

Show all work to receive full credit.

Comments for Math - Create Quadratic Equation

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Aug 12, 2012
Create Quadratic Equation
by: Staff


Answer:


Part I

To answer this question, you can use either the standard form of a quadratic function, or the vertex form of a quadratic function.


The standard form of a quadratic function (a parabola) is:

y = ax² + bx + c


The vertex form of a quadratic function (a parabola) is:

y = a(x – h)² + k


The quadratic function for a parabola in standard form can be converted to vertex form by completing the square.

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I am going to use the standard form of a quadratic function:

y = ax² + bx + c


The standard form of quadratic equation would be:

ax² + bx + c = 0


Solve for x using the quadratic formula:

x = (-b/2a) ± [√(b² - 4ac)/2a]


the axis of symmetry (the x coordinate of the vertex) of the parabola is:

x_axis_of_symmetry = (-b/2a)

Since the axis of symmetry must be x = 1:

(-b/2a) = x_axis_of_symmetry

(-b/2a) = 1


Choose any value for a, and then solve for b

Or

Choose any value for b, and then solve for a


As an example, since the parabola must open downward, a must be negative.

let a = -1

(a = -1 is an arbitrary choice. You can choose any value for “a”, as long as it is a negative value.)

(-b/2a) = 1

-b/[2*(-1)] = 1

-b/(-2) = 1

b/(2) = 1

[b/(2)]*(2) = 1*(2)

b*(2/2) = 1*2

b*(1) = 1*2

b = 2


y = ax² + bx + c

substituting the values a = -1 and b = 2

y = -1*x² + 2*x + c

y = -x² + 2x + c


you can now choose a value for c

Strictly speaking, you can choose any value for “c”. However, your problem statement asks that you compute the x-intercepts. “c” must be a positive value for the quadratic function to cross the x-axis. So I recommend you choose a positive value for “c”.

for example, let c = 5


using the values a = -1, b = 2, and c = 5

>>> The standard form of the final quadratic function is:

y = -x² + 2x + 5


>>> The standard form of the final quadratic equation is:

0 = -x² + 2x + 5


>>> The vertex form of the final quadratic function is:

y = -(x – 1)² + 6


>>> The vertex form of the final quadratic equation is:

0 = -(x – 1)² + 6




Open the following link to view a graph of y = -x² + 2x + 5



(1) If your browser is Firefox, click the following link to VIEW the graph; or if your browser is Chrome, Internet Explorer, Opera, or Safari (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page

http://www.solving-math-problems.com/images/parabola-axis-of-symmetry-equal-1.png

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Aug 12, 2012
Create Quadratic Equation
by: Staff

---------------------------------------------

Part II


Find the vertex:

Since the axis of symmetry = 1, x = 1 at the vertex

Use the quadratic function to calculate the y value

y = -x² + 2x + 5

substitute 1 for every x in the equation

y = -1² + 2*1 + 5

y = -1 + 2 + 5

y = 6

>>> the vertex is: (1,6)


---------------------------------------------
Find the domain:


Domain of the function

-∞ < x < ∞

---------------------------------------------
Find the range:


Range of the function

y ≤ 6

---------------------------------------------
Find the x-intercepts:

0 = ax² + bx + c

0 = -x² + 2x + 5

a = -1

b = 2

c = 5

x = (-b/2a) ± [√(b² - 4ac)/2a]

x = (-2/(2*(-1))) ± (√(2² - 4*(-1)*5)/(2*(-1))


x = 1 ± (-sqrt(6))

x₁ = 1 - sqrt(6), or -1.4494897427832

x₂ = 1 + sqrt(6), or 3.4494897427832

the x intercepts are:

x ∈ { 1 - sqrt(6), 1 + sqrt(6)}

or

x ∈ { -1.4494897427832, 3.4494897427832}


------------------------------------------

In summary:


Create your own quadratic equation:

1. that has an axis of symmetry of x = 1

2. whose graph opens down.


>>> The standard form of the final quadratic function is:

y = -x² + 2x + 5


>>> The vertex form of the final quadratic function is:

y = -(x – 1)² + 6



>>> the Vertex is: (1,6)

>>> the Domain of the function: -∞ < x < ∞

>>> the Range of the function: y ≤ 6

>>> the x intercepts:

x ∈ { 1 - sqrt(6), 1 + sqrt(6)}

or

x ∈ { -1.4494897427832, 3.4494897427832}



also see:

Algebra 1 (Honors)


Thanks for writing.

Staff
www.solving-math-problems.com



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