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Warm Up Graph: π¦= π₯ 2 +2π₯β3 STEPS FOR GRAPHING y = ax2 + bx + c
Step 1: Find and plot the vertex. The x βcoordinate of the vertex is π= βπ ππ Substitute this value for x in the equation and evaluate to find the y -coordinate of the vertex. Step 2: Draw the axis of symmetry. It is a vertical line through the vertex. Equation of axis of symmetry is x = # (x-coordinate of the vertex). Step 3: Make an x-y chart. Choose 2 (or more) values for x to the right or left of the line of symmetry. Plug them in the equation and solve for y. Step 4: Graph the points. Mirror the points on the other side of the line of symmetry. Draw a parabola through the points.
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5.2 Graphing Quadratic Functions in Vertex Form
11/13/13
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Equations of a lineβ¦ Slope intercept form: y = mx + b Ex: y = 3x + 2 Point β slope form: y - y1 = m(x β x1) Ex: y β 5 = 3(x β 1) Standard form: Ax + By = C Ex: 3x β 2y = 5
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Equations of a Parabolaβ¦
Standard Form: y = ax2 + bx + c Ex: y = 2x2 + 4x + 9 Vertex Form: y = a(x-h)2 + k
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When a is positive, the parabola opens up.
Vertex Form: a quadratic equation written in the form Y = a(x βh)2 + k (h, k) is the vertex Againβ¦ When a is positive, the parabola opens up. When a is negative, the parabola opens down. (h, k) Graphing Method: Step 1 Plot the vertex (h, k) from the equation Steps 2-4 : same as the standard form.
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Example 1 Graph ( )2 2 x β y = 1 + The function is in vertex form where , h 2, and k Because a is negative, the parabola opens down. = a 2 β y k + ( )2 h x STEP 1 vertex. ( ) h, k 2, 1 = STEP 2 Draw the axis of symmetry through the vertex. STEP 3 x y 1 -7 -1 When x = 0 π¦=β2 0β =β2 β =β2(4)+1 =β8+1=β7 When x = 1 π¦=β2 1β =β2 β =β2(1)+1 =β2+1=β1 STEP 4 Plot (0, -7) and (1, -1). Mirror them on the other side of axis of symmetry. Draw a smooth curve through the points.
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, h = 3, and k = -1. Because a is positive the parabola opens up. a =
Example 2 ( )2 3 x β y = 1 Graph , h = 3, and k = -1. Because a is positive the parabola opens up. a = 1 STEP 1 vertex. ( ) h, k 3, -1 = STEP 2 Draw the axis of symmetry through the vertex. STEP 3 x y 4 5 3 When x = 4 π¦= 4β3 2 β1 = (1) 2 β1 =0 When x = 5 π¦= 5β3 2 β1 = (2) 2 β1 =4β1 =3 STEP 4 Plot (4, 0) and (5, 3). Mirror them on the other side of axis of symmetry. Draw a smooth curve through the points.
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, h = -1, and k = 4. Because a is positive the parabola opens up. a =
Example 3 Graph 4 y = ( )2 1 x 2 + , h = -1, and k = 4. Because a is positive the parabola opens up. a = 2 STEP 1 vertex. ( ) h, k -1, 4 = STEP 2 Draw the axis of symmetry through the vertex. STEP 3 x y 1 6 12 When x = 0 π¦= = =2(1)+4 =2+4=π When x = 1 π¦= = =2(4)+4 =8+4=ππ STEP 4 Plot (0, 6) and (1, 12). Mirror them on the other side of axis of symmetry. Draw a smooth curve through the points.
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Maximum and Minimum Value of the Quadratic Eqn.
Vertex is the highest point, therefore the y-coordinate of the vertex is the maximum value. Vertex is the lowest point, therefore the y-coordinate of the vertex is the minimum value.
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Take y-coordinate as the minimum value of the function.
Find the Minimum or Maximum Value Tell whether the function has a minimum or maximum value. Then find the minimum or maximum value. ( )2 8 x y = 12 β 2 1 + Find the vertex: (h, k) is (-8, -12) Since a is positive, the parabola opens up and the vertex is the lowest point. Take y-coordinate as the minimum value of the function. ANSWER minimum; 12 β
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π¦=β π₯ 2 β2π₯+1 Tell whether the function has a minimum or maximum value. Then find the minimum or maximum value. π=β1, b=β2, c=1 π= βπ ππ π= β(βπ) π(βπ) = π βπ =βπ π¦= β1(β1) 2 β2 β1 +1=β1+2+1=2 Vertex (-1, 2) Since a is negative, the parabola opens down and the vertex is the highest point. Take y-coordinate as the maximum value of the function. ANSWER maximum; 2
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Homework WS 5.2
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