Do Now: Pass out calculators. Work on Practice EOC Week # 12 Write down your assignments for the week for a scholar dollar.

Objective: To graph parabolas using a table of values.

Quadratic Functions: A quadratic function is a nonlinear function that can be written in standard form: y = ax 2 + bx + c Every quadratic function has a U-shaped graph called a parabola.

Parent Quadratic Function: The most basic quadratic function in the family of quadratic functions is called the parent quadratic function: y = x 2 Vertex: The lowest or highest point on a parabola. Vertex of parent function: (0,0) Axis of Symmetry: The line that passes Through the vertex and divides the Parabola in half. Axis of Symmetry of parent function: x = 0

EXAMPLE 1 Graph y= ax 2 where a > 1 STEP 1 Make a table of values for y = 3x 2 x– 2– 1012 y12303 Plot the points from the table. STEP 2

EXAMPLE 1 STEP 3 Draw a smooth curve through the points. Compare the graphs of y = 3x 2 and y = x 2. Both graphs open up and have the same vertex, (0, 0), and axis of symmetry, x = 0. The graph of y = 3x 2 is narrower than the graph of y = x 2 because the graph of y = 3x 2 is a vertical stretch ( by a factor of 3) of the graph of y = x 2. STEP 4 Graph y= ax 2 where a > 1

EXAMPLE 2 Graph y = ax 2 where a < 1 Graph y = 1 4 – x2.x2. Compare the graph with the graph of y = x 2. STEP 1 Make a table of values for y = 1 4 – x2.x2. x – 4– 2024 y – 4– 10 – 4

EXAMPLE 2 STEP 2 Plot the points from the table. Draw a smooth curve through the points. STEP 3 Graph y = ax 2 where a < 1

EXAMPLE 2 STEP 4 Compare the graphs of y = 1 4 – x2.x2. and y = x 2. Both graphs have the same vertex (0, 0), and the same axis of symmetry, x = 0. However, the graph of 1 4 – x2x2 y = is wider than the graph of y = x 2 and it opens down. This is because the graph of 1 4 – x2x2 y = is a vertical shrink by a factor of 1 4 with a reflection in the x -axis of the graph of y = x 2. Graph y = ax 2 where a < 1

EXAMPLE 3 Graph y = x 2 + c Graph y = x Compare the graph with the graph of y = x 2. STEP 1 Make a table of values for y = x x– 2– 1012 y96569

EXAMPLE 3 Graph y = x 2 + c STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

EXAMPLE 3 Graph y = x 2 + c STEP 4 Compare the graphs of y = x and y = x 2. Both graphs open up and have the same axis of symmetry, x = 0. However, the vertex of the graph of y = x 2 + 5, (0, 5), is different than the vertex of the graph of y = x 2, (0, 0), because the graph of y = x is a vertical translation ( of 5 units up ) of the graph of y = x 2.

GUIDED PRACTICE for Examples 1, 2 and 3 Graph the function. Compare the graph with the graph of x y= –4x 2 ANSWER

GUIDED PRACTICE for Examples 1, 2 and 3 2. y = x ANSWER Graph the function. Compare the graph with the graph of x 2.

GUIDED PRACTICE for Examples 1, 2 and 3 3. y = x 2 +2 Graph the function. Compare the graph with the graph of x 2. ANSWER

EXAMPLE 4 Graph y = x 2 – 4. Compare the graph with the graph of y = x STEP 1 Make a table of values for y = x 2 – x– 4– 2024 y4 – 4–24 Graph y = ax 2 + c

EXAMPLE 4 Graph y = ax 2 + c STEP 2 Plot the points from the table. STEP 3 Draw a smooth curve through the points.

EXAMPLE 4 Graph y = ax 2 + c STEP 4 Compare the graphs of y = x 2 – 4 and y = x 2. Both graphs open up and have the same axis of symmetry, x = 0. However, the graph of y = x 2 – 4 is wider and has a lower vertex than the graph of y = x 2 because the graph of y = x 2 – 4 is a vertical shrink and a vertical translation of the graph of y = x

GUIDED PRACTICE for Example 4 Graph the function. Compare the graph with the graph of x y= 3x 2 – 6

GUIDED PRACTICE for Example 4 5. y= –5x 2 + 1

GUIDED PRACTICE for Example 4 6. y = x 2 –

EXAMPLE 5 Standardized Test Practice How would the graph of the function y = x be affected if the function were changed to y = x 2 + 2? A The graph would shift 2 units up. B The graph would shift 4 units up. C The graph would shift 4 units down. D The graph would shift 4 units to the left.

EXAMPLE 5 Standardized Test Practice SOLUTION The vertex of the graph of y = x is 6 units above the origin, or (0, 6). The vertex of the graph of y = x is 2 units above the origin, or (0, 2). Moving the vertex from (0, 6) to (0, 2) translates the graph 4 units down. ANSWER The correct answer is C. A BC D

Properties of a Quadratic Function y = ax 2 +bx + c Opens up a > 0 Opens down if a < 0 Shrink if a > 1 Wider if a < 1 Axis of Symmetry: x = -b/2a Vertex with an x-coordinate of x = -b/2a Has a y-intercept of c. So, the point (0,c) is on the parabola.

EXAMPLE 1 Find the axis of symmetry and the vertex Consider the function y = – 2x x – 7. a. Find the axis of symmetry of the graph of the function. b. Find the vertex of the graph of the function.

EXAMPLE 1 Find the axis of symmetry and the vertex SOLUTION 12 2(– 2) x = –x = – b 2a = = 3= 3 Substitute – 2 for a and 12 for b. Then simplify. For the function y = –2x x – 7, a =  2 and b = 12. a. b. The x- coordinate of the vertex is, or 3. b 2a – y = –2(3) (3) – 7 = 11 Substitute 3 for x. Then simplify. ANSWER The vertex is (3, 11).

EXAMPLE 2 Graph y = ax 2 + bx + c Graph y = 3x 2 – 6x + 2. Determine whether the parabola opens up or down. Because a > 0, the parabola opens up. STEP 1 STEP 2 = Find and draw the axis of symmetry: x = – b 2a2a – – 6– 6 2(3) =1. STEP 3 Find and plot the vertex.

EXAMPLE 2 To find the y- coordinate, substitute 1 for x in the function and simplify. y = 3(1) 2 – 6(1) + 2 = – 1 So, the vertex is (1, – 1). STEP 4 Plot two points. Choose two x- values less than the x- coordinate of the vertex. Then find the corresponding y- values. Graph y = ax 2 + b x + c The x- coordinate of the vertex is b 2a2a, or 1. –

EXAMPLE 2 Standardized Test Practice x 0 – 1 y 211 STEP 5 Reflect the points plotted in Step 4 in the axis of symmetry. STEP 6 Draw a parabola through the plotted points.

EXAMPLE 2 Graph y = ax 2 + b x + c

GUIDED PRACTICE for Examples 1 and 2 1. Find the axis of symmetry and vertex of the graph of the function y = x 2 – 2x – 3. ANSWER x = 1, (1, –4). 2. Graph the function y = 3x x – 1. Label the vertex and axis of symmetry. ANSWER

Minimum and Maximums The y-coordinate of the vertex is the minimum or maximum value.

EXAMPLE 3 Find the minimum or maximum value Tell whether the function f(x) = – 3x 2 – 12x + 10 has a minimum value or a maximum value. Then find the minimum or maximum value. SOLUTION Because a = – 3 and – 3 < 0, the parabola opens down and the function has a maximum value. To find the maximum value, find the vertex. x = – = – = – 2 b 2a2a – 12 2(– 3) f(–2) = – 3(–2) 2 – 12(–2) + 10 = 22 Substitute –2 for x. Then simplify. The x-coordinate is – b 2a2a

Find the minimum or maximum value EXAMPLE 3 ANSWER The maximum value of the function is f(– 2) = 22.

Find the minimum value of a function EXAMPLE 4 The suspension cables between the two towers of the Mackinac Bridge in Michigan form a parabola that can be modeled by the graph of y = x 2 – 0.37x where x and y are measured in feet. What is the height of the cable above the water at its lowest point? SUSPENSION BRIDGES

Find the minimum value of a function EXAMPLE 4 SOLUTION The lowest point of the cable is at the vertex of the parabola. Find the x-coordinate of the vertex. Use a = and b = – x = – = – ≈ 1910 b 2a2a – ( ) Use a calculator. Substitute 1910 for x in the equation to find the y-coordinate of the vertex. y ≈ (1910) 2 – 0.37(1910) ≈ 196

EXAMPLE 4 ANSWER The cable is about 196 feet above the water at its lowest point. Find the minimum value of a function

Exit Ticket Graph the Quadratic Equation below and label the following: vertex, max or min, axis of symmetry, and compare the parabola to its parent function. y = 2x 2 – 8x + 6. Draw a picture making your parabola into a real life example!