What is the vertex of the function f(c)=x^2+12x

Vertex:

(

6

,

−

38

)

x

-intercepts are:

(

6

+

√

38

,

0

)

and

(

6

−

√

38

,

0

)

y

-intercept is:

(

0

,

−

2

)

Explanation:

From this form of the parabola equation, i.e.

a

x

2

+

b

x

+

c

, the

x

-coordinate of the vertex is:

x

=

−

b

2

a

Here,

a

=

1

and

b

=

−

12

, and

c

is

−

2

therefore:

x

=

−

(

−

12

)

2

(

1

)

=

12

2

=

6

Now we plug that into the equation to find

y

:

y

=

6

2

−

12

(

6

)

−

2

=

36

−

72

−

2

=

36

−

74

=

−

38

Vertex:

(

6

,

−

38

)

To find the

x

-intercepts, you have to set

y

(the equation) equal to

0

and solve for

x

x

2

−

12

x

−

2

=

0

We cannot factor this. Let's use the quadratic formula:

x

=

−

b

±

√

b

2

−

4

a

c

2

a

x

=

−

(

−

12

)

±

√

(

−

12

)

2

−

4

(

1

)

(

−

2

)

2

(

1

)

x

=

12

±

√

144

+

8

2

=

12

±

√

152

2

=

12

±

√

2

2

⋅

38

2

=

12

±

2

√

38

2

=

2

(

6

±

√

38

)

2

=

2

(

6

±

√

38

)

2

x

=

6

±

√

38

Therefore, the

x

-intercepts are:

(

6

+

√

38

,

0

)

and

(

6

−

√

38

,

0

)

To find the

y

-intercept, you have to plug in

0

for

x

:

y

=

0

2

−

12

(

0

)

−

2

=

−

2

Therefore, the

y

-intercept is:

(

0

,

−

2

)

Step-by-step explanation:

What is the vertex of the function f(c)=x^2+12x

Step-by-step explanation:

one of each metals

step-by-step explanation: