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Oct 22 Quadratic Functions Homework: 4.2 #5, 17, 19, 27, 33, 39, 43, 45, 47, 53, 55 Quadratic functions define quadratic function. A quadratic function is a function defined by a quadratic polynomial, f ( x ) = a2 x 2 + a1 x + a0 where a2 , a1, a0 constants with a2 ! 0 or (more commonly) f ( x ) = ax 2 + bx + c where a, b, c constants with a ! 0 . E.g., y = !2x 2 + 3x !1. define quadratic-like functions. Some functions have a similar form to quadratic functions, e.g., f ( x ) = !2x + 3 x !1, g( x ) = !2x 4 + 3x 2 !1, and h ( x ) = !2e 2 x + 3e x !1. Each of these functions can be transformed into a quadratic function by using an appropriate substitution, e.g., in g we can substitute y = x 2 to get g( y ) = !2y 2 + 3y !1. We can now find the zeros of g( x ) = !2x 4 + 3x 2 !1 by first finding the zeros of g( y ) = !2y 2 + 3y !1, 1 1 and we find y = 1, y = . We back substitute for y to get x 2 = 1, x 2 = . 2 2 2 Solving for x, we get x = ±1, x = ± . 2 graphs of quadratic functions, parabola, vertex, axis of symmetry. Graphs of quadratic functions are curves called parabolas. Every parabola has two important features: an axis of symmetry and a vertex. The vertex is either a global maximum or minimum point and the axis of symmetry is a vertical line that passes through the vertex. The graph of every quadratic function also has a y-intercept and either zero, one or two x-intercepts. Find the equation of the parabola with a vertical axis of symmetry that passes through the points (1,0), (2,9), and (!3,4 ) . translations of y = x 2 , vertex quadratic form. Draw graph of y = x 2 . We have sketched the graph of y = !x 2 + 2x + 2 previously by converting the equation to y = !( x !1) + 3, where the vertex is (1,3) . This suggests 2 another quadratic form (called the vertex form), i.e., y = a( x ! h) 2 + k , where (h,k) is the vertex and x = h is the equation of the axis of symmetry. If a > 0 , the parabola opens upward; if a < 0 , it opens downward. The vertex form is useful for finding the vertex and the axis of symmetry easily. vertex form of quadratic functions. We can find general formulas for calculating the vertex and axis of symmetry for f (x) = ax 2 + bx + c by completing its square, i.e., f (x) = ax 2 + bx + c = b b b b b b2 a(x 2 + x) + c = a(x 2 + x + [ ]2 ) + c ! a[ ]2 = a(x + ) 2 + c ! . a a 2a 2a 2a 4a b b2 This is now in vertex form and the vertex is the point (! , c ! ) and 2a 4a b the axis of symmetry is x = ! . Find the equation whose 2a graph is a parabola with vertex ( 4,!3) , a vertical axis of symmetry, and passes through the point (5,!6) . factored form of quadratic functions, discriminant. The quadratic formula, !b ± b 2 ! 4ac x= , calculates zeroes of the function f ( x) = ax 2 + bx + c . 2a These zeroes (or roots of the equation ax 2 + bx + c = 0 ) may be real or complex. Let r1 and r2 represent the roots of the equation, then we can rewrite f(x) as f ( x) = a( x ! r1 )( x ! r2 ) . Early we stated that the roots, r1 and r2 , could be two distinct real roots, one repeated real root, or two complex roots. The discriminant, b 2 ! 4ac , will determine what type of roots we have. If b 2 ! 4ac > 0 , then we have two distinct real roots. If b 2 ! 4ac = 0 , one repeated real root. If b 2 ! 4ac < 0 , two complex roots. Why? Notice that complex roots always occur in conjugate pairs, i.e., b b 2 ! 4ac x =! ± i . Find the quadratic function whose graph has 2a 2a zeros at x = !5 and x = 3 and a y-intercept at (0, 30). Using the general form, f ( x) = ax 2 + bx + c substitute the points (!5,0), ( 3,0), and (0,30) for x and y. We obtain three functions: 25a ! 5b + c = 0 9a + 3b + c = 0 c = 30 Solving for A, B, and C, A = -2, B = -4, and C = 30. So, y = !2x 2 ! 4 x + 30 . Using the vertex form. We know that h lies on the axis of symmetry, which is the midpoint of the two zeros, so h = !1 . So, y = a( x +1) 2 + k . Again, after substituting our three points, we obtain 3 functions: 16a + k = 0 16a + k = 0 a + k = 30 Solving for a and k, we obtain a = !2 and k = 32 . So, y = !2( x +1) 2 + 32 . Using the factored form, f ( x) = a( x ! r1 )( x ! r2 ) . We know r1 and r2 . Therefore, y = a( x + 5)( x ! 3) . Substituting the point (0, 30), we obtain !15a = 30 , so a = -2 and y = !2( x + 5)( x ! 3) . Show that !2( x + 5)( x ! 3) = !2x 2 ! 4 x + 30 = !2( x +1) 2 + 32 . r1 + r2 compute vertex and axis of symmetry. The axis of symmetry is x = (the 2 average of the two roots). The y-value of the vertex can be found by r1 + r2 a substituting x = into f ( x) and calculating its value, ! (r1 ! r2 ) 2 . 2 4 r1 + r2 a Therefore, the vertex is ( ,! (r1 ! r2 ) 2 ) . Compute the vertex and axis 2 4 of symmetry of y = !2( x + 5)( x ! 3) . compute intercepts. To find the x-intercepts of the graph of any function f (if they exist), we solve the equation f ( x ) = 0 . To find the y-intercept, we calculate the value f (0) . Find the intercepts of y = !2x 2 ! 4 x + 30 . application. Find the equation of the tangent lines to the circle x 2 + y 2 = 10 from the point (1,!7) .