2. f(x) = x^2 + 2x + 1 vertex
axis
x-int
y-int
max/min
value
domain
range

For this problem we have 3 options.

1) We can use the completing the square the method to put the equations in Standard Vertex Form,

(

x

−

h

)

2

−

k

=

0

→

v

e

r

t

e

x

(

h

,

−

k

)

2) We can use the expression

(

−

b

2

a

,

f

(

−

b

2

a

)

)

where

a

and

b

are the coefficients found when the equation is in standard form,

a

x

2

+

b

x

+

c

=

0

3) We could just graph it.

Let do options 1 & 2.

Completing the Square:

x

2

−

2

x

−

1

=

0

x

2

−

2

x

=

1

(

−

2

2

)

2

=

(

−

1

)

2

=

1

, We will add

1

to both sides of the equation.

x

2

−

2

x

+

1

=

1

+

1

x

2

−

2

x

+

1

=

2

(

x

−

1

)

2

=

2

(

x

−

1

)

2

−

2

=

0

→

v

e

r

t

e

x

→

(

1

,

−

2

)

Solution

(

x

−

h

)

2

−

k

=

0

→

v

e

r

t

e

x

→

(

h

,

−

k

)

Using the expression

(

−

b

2

a

,

f

(

−

b

2

a

)

)

:

a

=

1

,

b

=

−

2

,

c

=

−

1

x

=

−

b

2

a

=

−

−

2

2

(

1

)

=

2

2

(

1

)

=

2

2

=

1

y

=

f

(

−

b

2

a

)

=

f

(

1

)

=

(

1

)

2

−

2

(

1

)

−

1

=

1

−

2

−

1

=

−

2

v

e

r

t

e

x

→

(

1

,

−

2

)

Solution

Notice that both methods produced the same results.

probably .5

i can on others if necessary.