logo for solving-math-problems.com
leftimage for solving-math-problems.com

Algebra 1 (Honors) // Quadratic Equations











































Create your own quadratic equation that has an axis of symmetry of x = 1 and whose graph opens down. Then find the vertex, domain, range, and x-intercepts. Show all work to receive full credit.

Comments for Algebra 1 (Honors) // Quadratic Equations

Click here to add your own comments

May 05, 2013
Quadratic Equation
by: Staff


Answer


Part I


The most convenient and direct way to control the shape and placement of a parabola on the x-y grid is to write the equation in the vertex form.

The vertex form of a parabola is:

Vertex form of a parabola





When the parabola is plotted, it looks like this:

Vertex form of a parabola - graph






When the coefficient "a" > 0, the parabola opens upward.

When the coefficient "a" < 0, the parabola opens downward.

Vertex form of a parabola - graph - coefficient 'a' is greater than 0





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

May 05, 2013
Quadratic Equation
by: Staff


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


Part II

Vertex form of a parabola - graph - coefficient 'a' is less than 0





Your Question:

Create your own quadratic equation that has an axis of symmetry of x = 1 and whose graph opens down.



The graph will open downward if a < 1.

I have arbitrarily chosen a = -3.

Vertex form of a parabola - selecting the coefficient 'a' so than the parabola opens downward





The axis of symmetry must be x = 1. Therefore h = 1.

Vertex form of a parabola - selecting the 'h' value so than the axis of symmetry is x = 1







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

May 05, 2013
Quadratic Equation
by: Staff


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


Part III



The value of k can be any number. I have arbitrarily chosen k = 12.

Vertex form of a parabola - selecting the 'k' value





The final equation:

Vertex form of a parabola - final equation - a = -3, h = 1, k = 12





Solving for the x-intercepts

set y = 0, and then solve for x

Vertex form of a parabola - final equation:  solve for x-intercepts





subtract 12 from each side of the equation

Subtract 12 from each side of the equation






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

May 05, 2013
Quadratic Equation
by: Staff


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


Part IV


divide each side of the equation by -3

Divide each side of the equation by -3





take the square root of each side of the equation

Take the square root of each side of the equation





1st intercept

Solve for first x-intercept





2nd intercept

Solve for second x-intercept








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

May 05, 2013
Quadratic Equation
by: Staff


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


Part V


Final Answer:

Equation:

Final equation for the parabola which opens downward with an axis of symmetry x = 1





Vertex:

Vertex of parabola is (1, 12)





Domain of the function:

Domain of parabolic function





Range of the function:

Range of parabolic function









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

May 05, 2013
Quadratic Equation
by: Staff


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


Part VI


x-intercepts:

x-intercepts of parabolic function





A graph for the parabola is shown below:

graph of parabola y = a(x - h)² + k






also see:

Create Quadratic Equation






Thanks for writing.

Staff
www.solving-math-problems.com


Click here to add your own comments

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.



Copyright © 2008-2015. All rights reserved. Solving-Math-Problems.com