Which equation has a graph that is a parabola with a vertex at (5, 3)?

a. y = (x – 5)2 + 3
b. y = (x + 5)2 + 3
c. y = (x – 3)2 + 5
d. y = (x + 3)2 + 5

we know that

The vertex form of a quadratic equation is equal to

where

(h,k) is the vertex of the parabola

Let's analyze each case to determine the solution

case a)

in this problem

using a graph tool

see the attached figure N

case b)

in this problem

using a graph tool

see the attached figure N

case c)

in this problem

using a graph tool

see the attached figure N

case d)

in this problem

using a graph tool

see the attached figure N

therefore

the answer is

The equation that has a graph that is a parabola with a vertex at is

case a)

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x