Our Parabola & Polynomial Goals -- February

Our Parabola & Polynomial Goals -- February On Tuesday’s Quiz 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

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 ). • Note: ‘h’ is (must be) subtracted from x. • also • If a>0 then it opens upwards, if a<0, then it opens downwards. • ‘a’ also tells us the “fatness”/ “skinniness” of the parabola

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 • 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 • 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 Can U Graph it?

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

Let’s do problems in book… page 270, #s 13, 14, 16, 17

If we’re given “h” and “k”, and a point on the parabola… We can 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

QUICK REVIEW • The steps to sketching a parabola: • Put in Standard Form • Identify the Vertex • Determine Direction • Determine where it crosses the y-axis • Sketch it. 1. f(x)=a(x-h)2+k  ____________ 2.Vertex  ( ___ , ___ ) 3.Direction  _________ 4.Y intercept  ( ____ , ____ ) Worksheet

POP QUIZ 50 Points • ID the vertex, the direction of the parabola & its y-intercept: • a) f(x) = 5(x – 2)2 + 4 • b) f(x) = –5(x – 1)2 + 3 • c) f(x) = – 2(x + 3)2 + 4 • d) f(x) = (x – 3)2 + 2 Vertex  ( ___ , ___ ) Direction  ___________ Y intercept  ( ____ , ____ ) ANSWER FORmat

POP QUIZ • Now Graph each of them. • a) f(x) = 5(x – 2)2 + 4 • b) f(x) = –5(x – 1)2 + 3 • c) f(x) = – 2(x + 3)2 + 4 • d) f(x) = (x – 3)2 + 2

What if NOT in Standard Form? Put in Std Form by Completing the Square!! When ‘a’ = 1 •  a(x2+bx)+c • Coefficient in front of x2 must be 1. • take half of the ‘b” term and square it, (b/2)2 . Add the result to the expression inside the parenthesis above and subtract it as well. • a(x2+bx+ (b/2)2 ) + c − (b/2)2 • This creates a perfect square trinomial (x+b/2)2 + ck • Putting into standard Quadratic Function gives: • (x – − b/2)2 + c , h=−b/2 and k=c Becomes the c in the standard formula

What if NOT in Standard Form? Put in Std Form by Completing the Square When ‘a’ not equal to 1 ! • Coefficient in front of x2 must be 1, so pull the ‘a’ out •  a(x2+bx)+c • take half of the ‘b” term and square it, (b/2)2 . Add the result to the expression inside the parenthesis and subtract it as well  a(x2+bx+ (b/2)2 − (b/2)2 ) + c • Simplify to create a perfect square trinomial • (x+b/2)2 + c − a(b/2)2 • Putting into standard Quadratic Function gives: • (x – − b/2)2 + c , h=−b/2 and k=c Becomes the c in the standard formula

What if NOT in Standard Form? Put in Std Form by Completing the Square When ‘a’ not equal to 1 ! Problems page 271,#s 29-34 • Coefficient in front of x2 must be 1, so pull the ‘a’ out •  a(x2+bx)+c • take half of the ‘b” term and square it, (b/2)2 . Add the result to the expression inside the parenthesis and subtract it as well  a(x2+bx+ (b/2)2 − (b/2)2 ) + c • Simplify to create a perfect square trinomial • (x+b/2)2 + c − a(b/2)2 • Putting into standard Quadratic Function gives: • (x – − b/2)2 + c , h=−b/2 and k=c Becomes the c in the standard formula

Real Life Applications-Gallery Walk page 272, #s 78-84 • Groups of four with these roles: • Solver – math problem solution leader • Recorder – easel pad leader • Speaker during Gallery Walk • Time Keeper • Time tables are on the white board ! • So is the format of the expected easel chart !

Simple Example… from class 1x2 – 2x 1[ 1x2–2x +(2/2)2 – (2/2)2]+ 0 1[x2–2x +(1)2] – 1(1)2+ 0 [x–1]2 – 1

Real Life Applications - Gallery Walk Example… Problem #78 –.008x2+1.8x + 1.5 –.008 [ 1x2–225x +(112.5)2 – (112.5)2]+ 1.5 –.008[x2–225x +(112.5)2 ] +(–.008) (– (112.5)2+ 1.5 –.008[x–112.5]2 + (–.008)(– (112.5)2+ 1.5 –.008[x–112.5]2 + 102.75

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

Our Parabola & Polynomial Goals -- February