Chapter 5 Notes - CP Algebra II Date Lesson 5.1 Notes...

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CP Algebra II Date: ______________ Lesson 5.1 Notes Graphing Quadratic Functions VOCABULARY: Quadratic function – has the form ax 2 + bx + c where a 0. Parabola – The U-shaped graph of a quadratic function. Vertex – The lowest or highest point on the graph of a quadratic function. Axis of Symmetry – The graph of a quadratic function is symmetric with a vertical line through the vertex. * Standard Form : y = ax 2 + bx + c * Vertex Form : y = a(x – h) 2 + k , Where the vetex is ( h , k ) and the axis of symmetry is x = h. * Intercept Form : y = ( x- p )( x = q ) , The x-intercepts are p and q. The axis of symmetry is halfway between ( q , 0 ) and ( p , 0 ). *For all three forms, the graph opens up if a > 0 and opens down if a < 0. The graph of a quadratic in Standard Form y = ax 2 + bx + c The parabola opens up if a > 0 and opens down if a < 0. The parabola is wider than the graph of y = x 2 if a < 1 and narrower than the graph of y = x 2 if a > 1. The x – coordinate of the vertex is b 2 a The axis of symmetry is the vertical line x = b 2 a
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Example 1) Graph a quadratic in Standard form: Graph y = x 2 – 4x + 3. The coefficients are a = 1, b = -4, and c = 3. Since a > 0, the parabola opens up. To find the x-coordinate of the vertex, substitute 1 for a and -4 for b in the formula b 2 a = ( 4) 2(1) 2 To find the y – coordinate of the vertex, substitute 2 for x in the original equation, and solve for y. y = x 2 4 x + 3 = (2) 2 4(2) + 3 = 1 The vertex is ( 2 , 1 ). Plot the vertex and find two points to the right or left of the vertex by substituting for x and solving for y. Then use symmetry to plot two more points. Draw the parabola with a smooth curve. Example 2) Graph a quadratic in Vertex form: Graph y = 2(x – 3) 2 – 4 Use the form y = a(x –h) 2 + k, where a = 2, h = 3, and k = -4. Since a > 0, the parabola opens up. Plot the vertex ( h , k ) = ( 3 , -4 ) Find two points to the right or left of the vertex by substituting for x and solving for y. Then use symmetry to plot two more points. Draw the parabola with a smooth curve.
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Example 3) Graph a quadratic in Intercept form: Graph y = 1 2 ( x 1)( x + 3) Use the intercept form y = a( x – p )( x – q ), where a = 1 2 , p = 1 and q = -3. The intercepts are ( 1 , 0 ) and ( -3 , 0). The axis of symmetry is 1 + ( 3) 2 = 1 The x-coordinate of the vertex is -1. To find the y-coordinate substitute for x and solve for y. y = 1 2 ( 1 1)( 1 + 3) = 2 . Plot the vertex ( -1 , 2 ) and draw the parabola. Example 4) Write the quadratic function in standard form. Write y = 2(x – 3 )(x + 8) Write the original equation y = 2(x – 3 )(x + 8). Multiply the two binomials by distributing or using the FOIL method: y = 2(x 2 + 5x -24). Use the distributive property to find y = 2x 2 + 10x - 48.
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Standard Form Intercept Form Vertex Form Equation y = ax 2 + bx + c y = a(x-p)(x-q) y = a(x-h) ² +k Axis of Symmetry x= -b/2a x = (p+q)/2 x = h Vertex (-b/2a, Plug x to find y) ((p+q)/2, Plug x to find y) (h,k) x intercepts & solving (P,0), (Q,0) get x alone algebraically "a" value b2 - 4ac = discriminant Discriminant tells how many solutions there are if > 0 there are 2 solutions if = 0 there is one solution if < 0 there are no real solutions; 2 imaginary solutions If a > 0 opens up, if a < 0 opens down, if a > 1 opens skinny, if 0 < a < 1 opens wide.
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