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:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation:

it is 10.6 according to my calculations