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

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 a)if the parabola opens upwards or downwards b)the “fatness”/ “skinniness” of the parabola

Substitute O for x (y intercept) and O for y (x intercept) e.g. f(x)= 3(x-1) = 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 e.g. f(x)= 3(x-1) f(x)= 3(0-1) f(x)= 3( – 1) 2 – 9  f(x)= 3 – 9  f(x)= – 6  ( 0, – 6) y intercept

using h, k and intercepts f(x)=5(x-3) f(x)=1(x+3) 2 – 4 Then graph these

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

Either rewrite into f(x)=a(x-h) 2 +k OR Memorize for ax 2 +bx+c, memorize: Vertex is [ –b/2a, f(–b/2a) ] Axis of symmetry is at x= –b/2a