Today in Pre c alculus

1. Today in Precalculus • Go over homework • Notes: (no handout) • Quadratic Functions • Homework

2. Quadratic Functions Standard form: f(x) = ax2 + bx + c Vertex form: f(x) = a (x – h)2 +k Vertex of the parabola (h, k) Axis of symmetry: x = h Roots (x-intercepts, zeros) found by factoring, quadratic formula or solving vertex form =0.

3. Quadratic Functions To convert from standard form to vertex form using algebra: Complete the square. Example: y = 2x2 + 16x + 30 y = 2(x2 + 8x) + 30 Factor a only from x2 and x y = 2(x2 + 8x + 16)-32 + 30 Divide new b by 2 and square it, must also multiply by a and subtract to keep equation true. y = 2(x +4)2 – 2 Factor and simplify Parabola with vertex (-4, -2) and axis of symmetry x = -4. x – intercepts (-3,0), (-5,0)

4. Practice Example: Use algebra to describe y = 3x2 – 18x – 3 y = 3(x2 – 6x) – 3 Factor a from x2 and x y = 3(x2 – 6x + 9) – 27 – 3 Divide new b by 2 and square it, must also multiply by a and subtract to keep equation true. y = 3(x – 3)2 – 30 Factor and simplify Parabola with vertex (3, -30) and axis of symmetry x = 3.

5. Formulas for h and k Example: y = 2x2 + 16x + 31 k = 31 – (2)(-4)2 or k = 2(-4)2 +16(-4) + 31 k = -1 Vertex: (-4, -1)

6. Practice Use the formulas to find the vertex of y = 3x2 + 5x – 4

7. Example Write the equation for a parabola with a vertex (-1,2) that goes through (6,100). start with vertex form: y = a(x + 1)2 + 2 substitute point 100 = a(6 + 1)2 + 2 98 = 49a a = 2 y = 2(x + 1)2 + 2

8. Homework • Pg. 182: 23-29odd (also find x-intercepts, may use your calculator), 35-41odd