Mrs. Rivas

Mrs. Rivas 𝒙 − 𝟓 𝒙 − 𝟔 𝒙 𝒙² −𝟓𝒙 𝒙 𝒙² −𝟔𝒙 − 𝟑 −𝟑𝒙 𝟏𝟓 − 𝟑 −𝟑𝒙 𝟏𝟖
Ida S. Baker H.S. b) 𝒙 𝟐 −𝟖𝒙+𝟏𝟓 a) − 𝒙 𝟐 +𝟗𝒙−𝟏𝟖 −( − ) 𝒙² 𝟗𝒙 𝟏𝟖 𝟏𝟓×𝟏 𝟓×𝟑 −𝟏𝟓×−𝟏 −𝟓×−𝟑 𝟏𝟖×𝟏 𝟗×𝟐 𝟔×𝟑 −𝟏𝟖×−𝟏 −𝟗×−𝟐 −𝟔×−𝟑 𝒙 2 −𝟓𝒙−𝟑𝒙+𝟏𝟓 −(𝒙 2 −𝟔𝒙−𝟑𝒙+𝟏𝟖) 𝒙 − 𝟓 𝒙 − 𝟔 𝒙 𝒙² −𝟓𝒙 𝒙 𝒙² −𝟔𝒙 − 𝟑 −𝟑𝒙 𝟏𝟓 − 𝟑 −𝟑𝒙 𝟏𝟖 ( )( ) (𝒙−𝟓)(𝒙−𝟑) 𝒙−𝟓 𝒙−𝟑 −( )( ) −(𝒙−𝟔)(𝒙−𝟑) 𝒙−𝟔 𝒙−𝟑

International Studies Charter School.
Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation Essential Question # 1: What is the vertex from of a quadratic function? 𝒚=𝒂 𝒙−𝒉 ²+𝒌 Answer:

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation Graphing a Parabola 1. Identify and graph the vertex. (h, k) 2. Identify and draw the axis of symmetry. x = h 3. Find and plot one points on one side of the axis of symmetry. 4. Plot the corresponding on the other side of the axis of symmetry. 5. Sketch the graph.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation Graph the function 𝒇 𝒙 = 𝟏 𝟐 𝒙 𝟐 . 𝒚=𝒂 𝒙−𝒉 ²+𝒌 Vertex (𝟎,𝟎) Axis-Symmetry. 𝒙=𝟎 𝒚= 𝟏 𝟐 (𝟐)² = 𝟏 𝟐 (𝟒) 𝒙=𝟐 =𝟐 (𝟐,𝟐) 𝒚= 𝟏 𝟐 (𝟒)² = 𝟏 𝟐 (𝟏𝟔) 𝒙=𝟒 =𝟖 (𝟒,𝟖)

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation 𝒚=𝒂 𝒙−𝒉 ²+𝒌 Vertex (𝟎,𝟎) Axis-Symmetry. 𝒙=𝟎 𝒚=− 𝟏 𝟑 (𝟑)² =− 𝟏 𝟑 (𝟗) 𝒙=𝟑 =−𝟑 (𝟑,−𝟑)

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation 𝒚=𝒂 𝒙−𝒉 ²+𝒌 Vertex (𝟎,−𝟓) Axis-Symmetry. 𝒙=𝟎 𝒙=𝟏 𝒚= 𝟏 2 −𝟓 =𝟏−𝟓 =−𝟒 (𝟏,−𝟒) 𝒙=𝟑 𝒚= 𝟑 2 −𝟓 =𝟗−𝟓 =𝟒 (𝟑,𝟒) Translation is 5 units down.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation 𝒚=𝒂 𝒙−𝒉 ²+𝒌 Vertex (𝟒,𝟎) Axis-Symmetry. 𝒙=𝟒 𝒙=𝟓 𝒚= 𝟓−𝟒 2 =(𝟏)² =𝟏 (𝟓,𝟏) 𝒙=𝟔 𝒚= 𝟔−𝟒 2 =(𝟐)² =𝟒 (𝟔,𝟒) Translation is 4 units right.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation 2. 𝒈 𝒙 =𝒙²+𝟑 3. 𝒉 𝒙 =(𝒙+𝟏)² Translation is 3 units up. Translation is 1 units left.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation 4. 𝒇 𝒙 =𝟑 𝒙−𝟒 ²−𝟐 5. 𝒇 𝒙 =−𝟐 𝒙+𝟏 𝟐 +𝟒 Translation is 1 units left and 4 units up. Translation is 4 units right and 2 units down.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation State weather the graph Reflects over the x-axis (𝒂 =−𝒏𝒖𝒎𝒃𝒆𝒓), Stretch (𝒂 > 𝟏) or Shrinks (𝟎 < 𝒂 < 𝟏). A) 𝑦= 𝑥 C) 𝑦=2 𝑥− E) 𝑦=− 𝑥− Since 𝒂 = −𝟐 then the graph opens down and it reflects over the 𝒙−𝒂𝒙𝒊𝒔 and shrinks. Since 𝒂 =+ 𝟏 then the graph opens up. Since 𝒂= +𝟏 then the graph opens up and the graph stretches. F) 𝑦= 𝑥+2 2 −1 B) 𝑦= − 𝑥 D) 𝑦=−2 𝑥− Since 𝒂 = −𝟐 then the graph opens down and it reflects over the 𝒙−𝒂𝒙𝒊𝒔 and stretches. Since 𝒂= +𝟏 then the graph opens up and the graph shrinks. Since 𝒂 = −𝟏 then the graph opens down and it reflects over the 𝒙−𝒂𝒙𝒊𝒔.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation Minimum and maximum value ** The minimum or maximum value is ALWAYS the 𝒚=𝒌.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation What is the is the minimum or maximum value of the following graphs. A) B) Vertex (−𝟒,𝟐) Vertex (−𝟏,−𝟑) Since the graph opens up, it has a minimum value = -3. Since the graph opens down, it has a maximum value = 2.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation Domain and Range [𝑲,∞) (−∞,𝒌] Vertex (𝒉,𝒌) ** The Domain (𝒉) is all the real numbers. (−∞,∞) ** The Range (𝒌) is all real numbers  (for minimum value) or  (for maximum value) than the value of 𝒌.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation What is the is the domain and range of the following graphs. A) B) Vertex (−𝟒,𝟐) Vertex (−𝟏,−𝟑) Domain (h) = (-∞, ∞). Domain (h) = (-∞, ∞). Range (k) = [-3, ∞). Range (k) = (-∞, 2].

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation What is the vertex, axis of symmetry, the maximum or minimum, the domain and the range and the transformation of the parent function? 𝒂 =−𝟑 Vertex (𝟒,−𝟐) Axis-Symmetry. 𝒙=𝟒 Since a = 1 and negative the graph opens down and stretch. Since the graph opens down we have a maximum value of −𝟐 and a reflection over the x-axis. Domain (h) = all the real numbers. (-∞, ∞) Range (k) = all the real numbers ≤−𝟐. (-∞, -2] Transformation is 4 units right and 2 units down.

Mrs. Rivas International Studies Charter School. Section 4-1 Quadratic Functions and Transformation What is the vertex, axis of symmetry, the maximum or minimum, the domain and the range and the transformation of the parent function? 𝒂 =𝟎.𝟑 Vertex (−𝟏,𝟒) Axis-Symmetry. 𝒙=−𝟏 Since 0 < a < 1 and Positive the graph opens up and shrink. Since the graph opens up we have a minimum value of 𝟒. Domain (h) = all the real numbers. (-∞, ∞) Range (k) = all the real numbers ≥𝟒. [4,∞) Transformation is 1 units left and 4 units up.

Mrs. Rivas The solution is −𝟐, 𝟒 𝟐𝒙+𝟑𝒚=−𝟏𝟔 𝟏𝟎𝒙+𝟏𝟓𝒚=−𝟖𝟎 𝟓𝒙−𝟏𝟎𝒚=𝟑𝟎
Ida S. Baker H.S. (𝟓) 𝟐𝒙+𝟑𝒚=−𝟏𝟔 𝟏𝟎𝒙+𝟏𝟓𝒚=−𝟖𝟎 (−𝟐) 𝟓𝒙−𝟏𝟎𝒚=𝟑𝟎 −𝟏𝟎𝒙+𝟐𝟎𝒚=−𝟔𝟎 𝟑𝟓𝒚=−𝟏𝟒𝟎 𝟑𝟓 𝟑𝟓 𝟐𝒙+𝟑𝒚=−𝟏𝟔 𝒚=−𝟒 𝟐𝒙+𝟑(−𝟒)=−𝟏𝟔 𝟐𝒙−𝟏𝟐=−𝟏𝟔 The solution is −𝟐, 𝟒 + 𝟏𝟐 + 𝟏𝟐 𝟐𝒙=−𝟒 𝟐 𝟐 𝒙=−𝟐

Mrs. Rivas International Studies Charter School. Section 4-2 Standard Form of a Quadratic Function 𝒇 𝒙 =𝒂𝒙²+𝒃𝒙+𝒄 y-intercept Step 1: Check 𝒂: ☻ If 𝒂 > 0 the quadratic functions opens up and the vertex represent the minimum point. ☻ If 𝒂< 0 the quadratic functions opens down and the vertex represent the maximum point. Step 2: Use −𝒃 𝟐𝒂 to find the vertex. Step 3: Substitute x into the function to obtain the y, which is the minimum or maximum value.

Mrs. Rivas International Studies Charter School. Section 4-2 Standard Form of a Quadratic Function Example: Graph 𝑦=𝑥²+2𝑥+1. What is the minimum value of the function. minimum value means y 𝑦=𝑥²+2𝑥+1 𝒇 𝒙 =𝒂𝒙²+𝒃𝒙+𝒄 𝒂= 𝟏> 𝟎 ☻ If 𝒂 > 0 the quadratic functions opens up and the vertex represent the minimum point. Step 2: Use −𝒃 𝟐𝒂 to find the vertex. 𝑦=(−𝟏)²+2(−𝟏)+1 𝒙= −𝒃 𝟐𝒂 = −𝟐 𝟐(𝟏) =−𝟏 𝑦=0 Step 3: Substitute x into the function to obtain the y, which is the minimum or maximum value. Vertex (-1, 0) which is the minimum point. Then minimum value is 0, since the minimum value is the y.

Mrs. Rivas International Studies Charter School. Section 4-2 Standard Form of a Quadratic Function Example: continue Graph Graph 𝑦=𝑥²+2𝑥+1. What is the minimum value of the function. y-intercept Vertex (-1,0). x 𝒚=𝒙²+𝟐𝒙+𝟏 (x, y) 𝟎 2 + 𝟒 𝟎 +𝟏 (0, 1) 1 𝟏 2 + 𝟒 𝟏 +𝟏 (1, 6) 2 𝟐 2 + 𝟒 𝟐 +𝟏 (2, 9)

Mrs. Rivas International Studies Charter School. Section 4-2 Standard Form of a Quadratic Function Graph and identify the, vertex, axis of symmetry, maximum or minimum value, and the range of 𝒚=𝒙²+𝟐𝒙+𝟑. y-intercept 𝒇 𝒙 =𝒂𝒙²+𝒃𝒙+𝒄 𝒙= −𝒃 𝟐𝒂 = −(𝟐) 𝟐(𝟏) =−𝟏 𝒚= −𝟏 ²+𝟐 −𝟏 +𝟑=𝟐 Vertex: (-1, 2) Axis-sym.: x =-1 Minimum.: y = 2 Range: all real numbers ≥ 2 𝒙 𝒇 𝒙 =𝒙²+𝟐𝒙+𝟑 𝒚 (𝒙, 𝒚) 1 𝟏 2 +𝟐 𝟏 +𝟑 6 (1, 6) 2 −𝟐 2 +𝟐 −𝟐 +𝟑 11 (2, 11)

Mrs. Rivas International Studies Charter School. Section 4-2 Standard Form of a Quadratic Function

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