1. Find the x -intercept and y -intercept 1.3x – 5y = 15 2.y = 2x + 7 ANSWER (5, 0); (0, –3) ANSWER (, 0) ; (0, 7) 7 2 –

3. Parabola Parent Function:

4. Graphing Quadratic Functions y = ax 2 + bx + c

5. Quadratic Functions The graph of a quadratic function is a parabola. A parabola can open up or down. If the parabola opens up, the lowest point is called the vertex. If the parabola opens down, the vertex is the highest point. NOTE: if the parabola opened left or right it would not be a function! Vertex minimum maximum

6. y = ax 2 + bx + c The parabola will open down when the a value is negative. The parabola will open up when the a value is positive. Standard Form The standard form of a quadratic function is a > 0 a < 0

7. y x Line of Symmetry Parabolas have a symmetric property to them. If we drew a line down the middle of the parabola, we could fold the parabola in half. We call this line the line of symmetry. The line of symmetry ALWAYS passes through the vertex. Or, if we graphed one side of the parabola, we could “fold” (or REFLECT) it over, the line of symmetry to graph the other side. Line of Symmetry

8. Find the line of symmetry of y = 3x 2 – 18x + 7 Finding the Line of Symmetry When a quadratic function is in standard form The equation of the line of symmetry is y = ax 2 + bx + c, For example… Using the formula… This is best read as … the opposite of b divided by the quantity of 2 times a. Thus, the line of symmetry is x = 3.

9. Finding the Vertex We know the line of symmetry always goes through the vertex. Thus, the line of symmetry gives us the x coordinate of the vertex. To find the y coordinate of the vertex, we need to plug the x value into the original equation. STEP 1: Find the line of symmetry STEP 2: Plug the x value into the original equation to find the y value. y = –2x 2 + 8x –3 y = –2(2) 2 + 8(2) –3 y = –2(4)+ 8(2) –3 y = –8+ 16 –3 y = 5 Therefore, the vertex is (2, 5)

10. A Quadratic Function in Standard Form The standard form of a quadratic function is given by y = ax 2 + bx + c There are 3 steps to graphing a parabola in standard form. STEP 1: Find the line of symmetry STEP 2: Find the vertex STEP 3: Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve. Plug in the line of symmetry (x value) to obtain the y value of the vertex. MAKE A TABLE using x values close to the line of symmetry. USE the equation

11. STEP 1: Find the line of symmetry Let's Graph ONE! Try … y = 2x 2 – 4x – 1 A Quadratic Function in Standard Form Thus the line of symmetry is x = 1

12. Let's Graph ONE! Try … y = 2x 2 – 4x – 1 STEP 2: Find the vertex A Quadratic Function in Standard Form Thus the vertex is (1,–3). Since the x value of the vertex is given by the line of symmetry, we need to plug in x = 1 to find the y value of the vertex. *Since a is positive, we know the parabola opens up

13. 5 –1 Let's Graph ONE! Try … y = 2x 2 – 4x – 1 STEP 3: Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve. A Quadratic Function in Standard Form 3 2 yx

14. Warm-Up Exercises EXAMPLE 4 Find the minimum or maximum value Tell whether the function y = 3x 2 – 18x + 20 has a minimum value or a maximum value. Then find the minimum or maximum value. SOLUTION Because a > 0, the function has a minimum value. To find it, calculate the coordinates of the vertex. x = – b2a b2a = – (– 18) 2a = 3 y = 3(3) 2 – 18(3) + 20 = –7 ANSWER The minimum value is y = –7.

15. Warm-Up Exercises GUIDED PRACTICE for Examples 4 and 5 7. Find the y value of the vertex and tell whether It’s a minimum or maximum y = 4x 2 + 16x – 3. x = – b2a b2a = – 16 2(4) = – 16 8 = -2 y = 4(-2) 2 + 16(-2) – 3 y = 16 – 32 – 3 y = -19 -19 minimum

16. Classwork Assignment: WS 4.1 (1-21 odd, 22-26 all)