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

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Term Polynomials Degree Leading coefficient 2x-2

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

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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.

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

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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

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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.

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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.

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

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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.

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

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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)

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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 x Finding the Vertex -100/ 2(-1) (50,2500)

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

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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 t+58.8, where s is in meters. When does the object strike the ground?

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0=-4.9t t+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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