Which is the standard form of the equation of the parabola that has a vertex of (3,1) and a directrix of x= -2?

(y-1)^2=20(x-3)

Step-by-step explanation:

The parabola is sideways so the squared part will be on the y.

The equation will be using is (y-k)^2=4p(x-h)

where (h,k) is the vertex.

p tells us which way the parabola opens by the sign.

p also tells us how far the directrix is from the vertex or the vertex from the focus.

Anyways if you graph the vertex and the directrix you will see the parabola opens to the right so p is going to be positive.

Now since the directrix is at x=-2 and the vertex is at (3,1) then p=3-(-2)=5.  5 means that the directrix and vertex are 5 units apart.

The vertex was given as (3,1) so h=3 and k=1.

Let's plug this information in:

(y-1)^2=4*5(x-3)

Let's simplify just a little.

(y-1)^2=20(x-3)

i believe that the slope of the line is -5

step-by-step explanation:

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !