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Quadratic Functions

Chapter 2 Section 1. Quadratic Functions. By: Brooke Tellinghuisen Kelli Peters Austin Steinkamp. Vocabulary. Term. 2x-2. Polynomials. Degree. Leading coefficient. Definition Of Polynomial function. Example of Polynomial Functions Polynomials are classified by degree.

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Quadratic Functions

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

  2. Vocabulary Term 2x-2 Polynomials Degree Leadingcoefficient

  3. Definition Of Polynomial function • Example of Polynomial Functions • Polynomials are classified by degree. • Formula of Polynomial Function

  4. Definition of Quadratic function 2nd 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.

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

  6. Quadratics General form y=ax2+bx+ c Vertex Form y= a(x-h)2+k Factored form y=(x-r1)(x-r2) Vertex Vertex (h,k) Vertex: standard form or vertex form Roots: Standard Form or factored form Roots: Quadratic formula or factored form Roots: x= r1,r2

  7. 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.

  8. 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.

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

  10. Practice problem answer 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.

  11. Practice Problem 2 • Find the vertex and x-intercepts of the equation f(x)=x2-5

  12. Answer to practice problem 2 • Take (x2-5) and set equal to zero • x2-5=0 To find your vertex +5 =+5 x2 =5 Use the formula to find your vertex Plug 0 back into the equation and solve. Those would be your x-intercepts Your answer is (0, -5)

  13. AreaExample • 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 X-100=x(-x) -x2+100x Finding the Vertex -100/ 2(-1) (50,2500) 2x+2L=200 -2x -2x 2L=200-2x 2L/2=200-2x/2 L=100-x X X L

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

  15. Projectile motion problem • 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)=-4.9t2+19.6t+58.8, where s is in meters. • When does the object strike the ground?

  16. Answer to Projectile motion problem • 0=-4.9t2+19.6t+58.8 • 0=t2-4t-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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