2.1Quadratic Functions

Use the graphing calculator and graph y = x2

y = 2x2

y = 2

3

1x

The quadratic function f(x) = a(x – h)2 + kis said to be in standard form. Vertex (h,k)If a > 0, opens up. If a < 0, opens down.

Ex. Write the quadratic function in standardform, find the vertex, identify the x-int.’sand sketch. f(x) = 2x2 + 8x + 7

f(x) = 2x2 + 8x + 7 Factor out a 2 from the x’s

f(x) = 2(x2 + 4x ) + 7 Complete the square+ 4 - 8

f(x) = 2(x + 2)2 - 1 V( , ) Up or Down

Take the original and set = 0 to find the x-int.

0 = 2x2 + 8x + 7 Use quad. formula on calc.

x = -1.2928, -2.7071 Now sketch it.

-2 -1

(-2,-1)

Ex. Sketch the graph of f(x) = -x2 + 6x - 8

f(x) = -x2 + 6x - 8 Factor -1 out of the x terms

f(x) = -(x2 – 6x ) - 8 Complete the square+ 9 + 9

f(x) = -(x – 3)2 + 1 V(3, 1) Down

To find a vertex of the quadratic f(x) = ax2 + bx + c, (without putting the equation into standard form), evaluate by letting

a

bx

2

−=

Ex. P = .0014x2 - .1529x + 5.855

6.54)0014(.2

1529.

2≈

−−=−=

ab

xy = 1.68

Find the the equations of a parabola that cups up and down given the x-intercepts (1, 0) and (-3, 0).

First, write the x-intercepts as factors.

y = (x - 1)(x + 3)

The parabola that cups up is y = x2 + 2x - 3

The parabola that cups down requires a negativein front of the quadratic.

y = -(x2 + 2x - 3)or y = -x2 - 2x + 3

Given the vertex and point on a given parabola, find the standard equation of the parabola.

V(4, -1); point (2, 3)

Fill in the points in the standard parabola formy = a(x - h)2 + k and solve for a.

3 = a(2 - 4)2 + (-1)

3 = 4a -14 = 4a1 = a

The answer isy = 1(x - 4)2 - 1