0, the range will be y > k. If a < 0, the range will be y < k The h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0). The k represents a vertical shift (how far up, or down, the graph has shifted from y = 0). (We’ll talk more about these “shifts” later.)"> 0, the range will be y > k. If a < 0, the range will be y < k The h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0). The k represents a vertical shift (how far up, or down, the graph has shifted from y = 0). (We’ll talk more about these “shifts” later.)">
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Vertex Form of A Quadratic Function
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y = a(x – h) 2 + k The vertex form of a quadratic function is given by f (x) = a(x - h) 2 + k where (h, k) is the vertex of the parabola.
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y = a(x – h) 2 + k When written in "vertex form": (h, k) is the vertex of the parabola. x = h is the axis of symmetry. k is the minimum or the maximum of the parabola. If a > 0, the range will be y > k. If a < 0, the range will be y < k The h represents a horizontal shift (how far left, or right, the graph has shifted from x = 0). The k represents a vertical shift (how far up, or down, the graph has shifted from y = 0). (We’ll talk more about these “shifts” later.)
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Notice that the h value is subtracted in this form, and that the k value is added. If the equation is y = 2(x - 1) 2 + 5, the value of h is 1, and k is 5. (1, 5) If the equation is y = 3(x + 4) 2 - 6, the value of h is -4, and k is -6. (-4, -6)
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Graphing Equations in Vertex Form 1. Start with the function in vertex form: y = a(x - h) 2 + k 2. Pull out the values for h and k. If necessary, rewrite the function so you can clearly see the h and k values. (h, k) is the vertex of the parabola. Plot the vertex. 3. The line x = h is the axis of symmetry. Draw the axis of symmetry. y = 3(x – 2)2 – 4 y = 3(x - 2) 2 + (-4) h = 2; k = -4 Vertex: (2, -4) x = 2 is the axis of symmetry
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Graphing (cont) 4. Find two or three points on one side of the axis of symmetry, by substituting your chosen x-values into the equation. For this problem, we chose (to the left of the axis of symmetry): x = 1; y = 3(1 - 2) 2 - 4 = -1 x = 0; y = 3(0 - 2) 2 - 4 = 8 Plot (1, -1) and (0,8)
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Graphing (cont) 5. Plot the mirror images of these points across the axis of symmetry, or plot new points on the right side. Draw the parabola. Remember, when drawing the parabola to avoid "connecting the dots" with straight line segments. A parabola is curved, not straight, as its slope is not constant.
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Write the function described in vertex form: A parabola has its vertex at (0, 5) and passes through the point (-3, -13). Plug the vertex into y = a(x - h) 2 + k y = a(x - 0) 2 + 5 Now plug in the point (-3, -13) into x and y respectively: -13 = a(-3 - 0) 2 + 5 Simplify and solve for “a”. -13 = a(-3) 2 + 5
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Write the function (cont) -13 = a(-3) 2 + 5 -13 = 9a + 5 Subtract 5 from both sides -18 = 9a Divide both sides by 9 a = -2 Plug the vertex and “a” into y = a(x - h) 2 + k y = -2(x – 0) 2 + 5 or y = -2(x) 2 + 5
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Let’s Try Another: A parabola has its vertex at (-1, 2). It passes through the point (1, 6) y = a(x + 1) 2 + 2 6 = a(1 + 1) 2 + 2 6 = a(2) 2 + 2 6 = 4a + 2 4 = 4a a = 1 y = (x + 1) 2 + 2
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One more… A parabola has its vertex at (-1, -4). It passes through the point (-3, 8). y = a(x + 1) 2 – 4 8 = a(-3 + 1) 2 – 4 8 = a(-2) 2 – 4 8 = 4a – 4 12 = 4a a = 3 y = 3(x + 1) 2 - 4
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