Quadratic Functions. A quadratic function is of the form; f(x) = ax 2 + bx + c, a ≠ 0 1. The graph of f is a parabola, which is concave upward if a >

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Presentation transcript:

Quadratic Functions

A quadratic function is of the form; f(x) = ax 2 + bx + c, a ≠ 0 1. The graph of f is a parabola, which is concave upward if a > 0 and concave upward if a < 0 2. The graph intersects the y-axis at (0,c) 3. If exists x 1 such that f(x 1 ) = 0, then (x 1, 0) is a point of intersection of f with the x-axis. 4. If f(x) has no zeros, then the graph of f does not intersects the x-axis. And so: a. The graph is above the x-axis if a > 0 b. The graph is below the x-axis if a < 0 5. If, by completing the square, f(x) is written as: f(x)= m(x-k) 2 + n. Then the vertex of the parabola is (k,f ( k ) )=(k,n)

Example (1) Graph f(x)= -2x 2 + 5x The graph is concave downward. Why? 2. The intersection with the y-axis = ( 0, f (0) ) = (0,-2) 3. f(x)= - (2x 2 - 5x + 2) = - (2x-1)(x-2) The intersections with the x-axis (2,0) and (1/2, 0 ) 4. f(x)=-2(x 2 - 5/2 x ) + 2 = -2 [(x- 5/4 ) 2 – 25/16 ]-2 = -2 (x- 5/4 ) /8 - 2 = -2 (x- 5/4 ) 2 + 9/8 The vertex of the parabola is (5/4, 9/8 )

Example (2) Graph f(x)= 2x 2 - 4x The graph is concave upward. Why? 2. The intersection with the y-axis = ( 0, f (0) ) = (0,1) 3. f(x)= 2x 2 - 4x + 1 The intersections with the x-axis: (1+ √2/2, 0 ) and (1- √2/2, 0 ) 4. f(x)=2(x x ) + 1 = 2 [(x- 1 ) 2 – 1 ] + 1 = 2 (x- 1 ) = 2 (x- 1 ) The vertex of the parabola is (1, -1 )

Example (3) Graph f(x)= x 2 - 6x The graph is concave upward. Why? 2. The intersection with the y-axis = ( 0, f (0) ) = (0,7) 3. f(x)= x 2 - 6x + 7 The intersections with the x-axis: (3+ √2, 0 ) and (3- √2, 0 ) 4. f(x)=x x + 7 = (x- 3 ) 2 – = (x- 3 ) 2 – 2 The vertex of the parabola is (3, - 2 )

Example (3) Graph f(x)= 2x 2 + 2x The graph is concave upward. Why? 2. The intersection with the y-axis = ( 0, f (0) ) = (0,3) 3. No intersections with the x-axis: Why? 4. f(x)=2[x 2 +x] + 3 = 2[(x+½) 2 –¼] + 3 = 2(x+½) 2 –½ + 3 = 2(x+½) 2 + 5/2 The vertex of the parabola is ( - ½, 5/2 )

Example (3) Graph f(x)= -x 2 - 4x The graph is concave downward. Why? 2. The intersection with the y-axis = ( 0, f (0) ) = (0,12) 3. f(x)= -x 2 - 4x + 12=-(x 2 +4x - 12)=-(x+6)(x-2) The intersections with the x-axis: (2, 0 ) and (-6, 0 ) 4. f(x)= -(x 2 +4x ) + 12=-[(x+ 2 ) 2 – 4] + 12 = -(x+2) The vertex of the parabola is (-2, 16 )

Question Do it now! Do not see the next slide before you do it! Graph the function: f(x)= x 2 + 2x - 3

Solution f(x)= x 2 +2x The graph is concave upward. Why? 2. The intersection with the y-axis = ( 0, f (0) ) = (0,-3) 3. f(x)= x 2 + 2x - 3=(x+3)(x-1) The intersections with the x-axis: (-3, 0 ) and (1, 0 ) 4. f(x)= x 2 +2x – 3 = = (x+1) 2 – 1 – 3 = (x+1) The vertex of the parabola is (-1, -4 )