# By: Brooke Tellinghuisen Kelli Peters Austin Steinkamp Chapter 2 Section 1.

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By: Brooke Tellinghuisen Kelli Peters Austin Steinkamp Chapter 2 Section 1

Term Polynomials Degree Leading coefficient 2x-2

Example of Polynomial Functions Polynomials are classified by degree. Formula of Polynomial Function

2 nd degree polynomials functions are called quadratic functions. Example of Quadratic Functions Formula of Quadratic Function NOTE: a, b, and c are real numbers with a 0.

Used to find zeros (roots) in a quadratic function.

General form y=ax 2 +bx+ c Vertex Form y= a(x-h) 2 +k Factored form y=(x-r 1 )(x-r 2 ) Vertex (h,k) Vertex Roots: Standard Form or factored form Roots: Quadratic formula or factored form Vertex: standard form or vertex form Roots: x= r 1,r 2

The graph for a quadratic function is a “U”-shaped graph, called a parabola. If the leading coefficient is positive, the graph opens upward. If the leading coefficient is negative, the graph opens downward.

The point where the axis intersects the parabola is the vertex. If a > 0, the vertex is the point with the minimum y- value on the graph. If a < 0, the vertex is the point with the maximum y- value on the graph.

 F(x)=(x-2)^2  Tell what direction the graph moves and if it opens up or down.

Since the 2 is connected with the x in the parentheses the graph moves the opposite way of what u think it would. Since it’s a subtraction problem it moved to the right.

 Find the vertex and x-intercepts of the equation f(x)=x 2 -5

 Take (x 2 -5) and set equal to zero  x 2 -5=0 +5 =+5 x 2 =5 To find your vertex Those would be your x-intercepts Use the formula to find your vertex Plug 0 back into the equation and solve. Your answer is (0, -5)

 Area Problem A(x)=width x length A farmer has 200 yards of fencing. Write the area as a function of x, if the farmer encloses a rectangular area letting the width equal to x. What is my maximum area? What are my zeros? Do the zeros match common sense? L XX L 2x+2L=200 -2x 2L=200-2x 2L/2=200-2x/2 L=100-x X-100=x(-x) -x 2 +100x Finding the Vertex -100/ 2(-1) (50,2500)

A function of height that depends on time. a = acceleration of gravity. b = initial velocity in which object is thrown. c = initial height.

 An object is launched at 19.6 meters per second from a 58.8 meter tall platform.  The equation for the objects highest s at time t seconds after launch is s(t)=-4.9t 2 +19.6t+58.8, where s is in meters.  When does the object strike the ground?

 0=-4.9t 2 +19.6t+58.8  0=t 2 -4t-12  0=(t-6)(t+2)  So T=6 and -2. The answer cant be negative so the object hit the ground at 6 seconds after the launch.

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