Step 4: Factor the expression inside the parentheses. x 2 ( y 1)2 3 Step 5: Solve the equation for x by multiplying by 2. 2 ax 2 b 2( y 1)2 3 x 2( y 1)2 6 YOU TRY 5 Rewrite x y2 6y 1 in the form x a(y k)2 h by completing the square. 6 Find the Vertex of the Graph of x ay2 by c Using y b 2a , and Graph the Equation Graph x y2 2y 5. Find the vertex using the vertex formula. Find the x- and y-intercepts and the domain and range. Solution Since this equation is solved for x and is quadratic in y, it opens in the x-direction. a 1, so it opens to the right. Use the vertex formula to fi nd the y-coordinate of the vertex. y b 2a y 2 2(1) 1 a 1, b 2 Substitute y 1 into x y2 2y 5 to fi nd the x-coordinate of the vertex. x (1)2 2(1) 5 x 1 2 5 4 The vertex is (4, 1). Because the vertex is (4, 1) and the parabola opens to the right, the graph has no y-intercepts. To fi nd the x-intercept, let y 0 and solve for x. x y2 2y 5 x (0)2 2(0) 5 x 5 The x-intercept is (5, 0). Find another point on the parabola by choosing a value for y that is close to the y-coordinate of the vertex. Let y 1. Find x. x (1)2 2(1) 5 x 1 2 5 8 Another point on the parabola is (8, 1). Use the axis of symmetry to locate the additional points (5, 2) and (8, 3). The domain is 4, q), and the range is (q, q) EXAMPLE 6 x y 1 5 5 V 1 5 2 5 1 SECTION 10.6 Applications of Qu www.mhhe.com/messersmith adratic Functions and Graphing Other Parabolas 675
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