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Mastering Quadratic Functions and Polynomial Division: Goals for February

In this February learning material, students will master critical concepts related to quadratic functions and polynomial division. They will be able to identify the vertex, axis of symmetry, x and y intercepts of a given quadratic function in standard form. Students will learn to sketch parabolas and rewrite equations into standard form. Additionally, they will perform long and synthetic division on polynomials. With practical examples and problem-solving exercises, this content prepares students to graph parabolas and understand polynomial behavior effectively.

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Mastering Quadratic Functions and Polynomial Division: Goals for February

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  1. Our Parabola & Polynomial Goals -- February Material on Quiz and Exam Student will be able to: • If given Quadratic Function in Standard Form: • ID Vertex, Axis of Symmetry, x and y intercepts • Sketch Parabola • Rewrite equation into Quadratic Fcn‘s Std Form • ID LH and RH Behavior of Polynomials • Perform Long Division of Polynomials • Perform Synthetic Division of Polynomials

  2. Standard Form for a Quadratic Function (Parabola) f(x)=a(x-h)2+k • hAxis of symmetry, its a vertical line at x=h • kVertex at ( h, k ) • … leaving the “a”, which tells us • if the parabola opens upwards or downwards • the “fatness”/ “skinniness” of the parabola

  3. How would we find the x and y intercepts? Substitute O for x (y intercept) and O for y (x intercept) • e.g. f(x)= 3(x-1)2 - 9 • f(x)= 3(0-1)2 - 9 • f(x)= 3( – 1)2 – 9 • f(x)= 3 – 9 • f(x)= – 6 • ( 0, – 6) • y intercept • e.g. f(x)= 3(x-1)2 - 9 • 0= 3(x – 1)2 – 9 • 9=3(x – 1)2 • 9/3=(x – 1)2 • 3=(x – 1)2 • ±√3 = x – 1 • (±√3 +1, 0) • x intercept Can U Graph it?

  4. Sketch it… using h, k and intercepts #s 1 – 8, on page 270 f(x)=5(x-3)2 + 4 f(x)=1(x+3)2 – 4 Then graph these

  5. Let’s do problems in book… page 270, #s 13 – 18,

  6. If we’re given “h” and “k”, and a point on the parabola… We write its function, f(x) • e.g., if h=3, and k=5 and the parabola passes thru (0,0)… f(x)=a(x-h) 2+k • f(x)=a(x-3)2+5 • We can find a by substitution … • 0=a(0-3)2 +5 • –5=9a • – 5/9=a • f(x)= – 5/9(x-3)2+5

  7. What if it is written in the standard form of a quadratic equation… Either rewrite into f(x)=a(x-h)2+kOR Memorize • for ax2+bx+c, memorize: • Vertex is [ –b/2a, f(–b/2a) ] • Axis of symmetry is at x= –b/2a

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