Answered

how to Plot the graph of the parabola (x-1)² = -16 (y-4) on the cartesian plane​

Sagot :

Ophelia1107viz

Answer:

The general "vertex form" for a parabola (in standard position) is

XXX

y

=

m

(

x

−

a

2

)

+

b

with vertex at

(

a

,

b

)

Notice that the given equation

XXX

y

=

(

x

+

1

)

2

−

4

is almost in this form, and we could rewrite it as

XXX

y

=

1

(

x

−

(

−

1

)

)

2

+

(

−

4

)

with vertex at

(

−

1

,

−

4

)

The

y

-intercept is the value of

y

when

x

=

0

and using the given equation:

XXX

y

x

=

0

=

(

0

+

1

)

2

−

4

=

−

3

So

(

0

,

−

3

)

is a point on the parabola.

Note that the axis of symmetry (for a parabola in standard position) is a vertical line (i.e.

x

=

some constant) through the vertex;

so in this case the axis of symmetry is

x

=

−

1

.

If the axis of symmetry is

x

=

−

1

and

(

0

,

−

3

)

is a point on the parabola,

since

(

0

,

−

3

)

is a point

1

unit to the right of the vertical line

x

=

−

1

then there is another point

1

unit to the left of

x

=

−

1

with the same

y

coordinate, namely

(

−

2

,

−

3

)

The three points

(

−

1

,

−

4

)

,

(

0

,

−

3

)

,

and

(

−

2

,

−

3

)

should be sufficient to sketch the parabola (although, if you like, you could solve the given equation for the

x

−

∫

e

r

c

e

p

t

v

a

l

u

e

s

a

s

w

e

l

l

b

y

s

e

∈

g

y=0# in the original equation):