Question

Type the equation of the following graph in the format x-a=(b/c)(y/d)^e

Answer 1

@dpaInc @Mertsj

Answer 2

@jim_thompson5910

Answer 3

This is an equation of a parabola.


It goes through the points (1,2), (3,6) and (3,-2)


So for (1,2), this means x = 1 and y = 2. So we can plug them into the equation ay^2+by+c = x to get


ay^2+by+c = x


a(2)^2+b(2)+c = 1


4a+2b+c = 1

-------------------

We can then plug in x = 3 and y = 6 to get


ay^2+by+c = x


a(6)^2+b(6)+c = 3


36a+6b+c = 3

-------------------

We can then plug in x = 3 and y = -2 to get


ay^2+by+c = x


a(6)^2+b(-2)+c = 3


4a-2b+c = 3

---------------------------------------------------

So we have three equations with 3 unknowns

4a+2b+c = 1
36a+6b+c = 3
4a-2b+c = 3


Solve this system using either elimination, substitution, row reduction, or calculator to get


a = 1/8, b = -1/2, c = 3/2


So the equation of the parabola is (1/8)y^2-(1/2)y+3/2 = x


Now let's convert it to the form x-a=(b/c)(y-d)^e


Note: I'm assuming you meant "y-d" instead of "y/d"


(1/8)y^2-(1/2)y+3/2 = x


(1/8)y^2-(1/2)y = x - 3/2


x - 3/2 = (1/8)y^2-(1/2)y


x - 3/2 = (1/8)(y^2-4y)


x - 3/2 = (1/8)(y^2-4y+4-4)


x - 3/2 = (1/8)((y^2-4y+4)-4)


x - 3/2 = (1/8)(y-2)^2+(1/8)(-4)


x - 3/2 = (1/8)(y-2)^2-1/2


x - 3/2+1/2 = (1/8)(y-2)^2


x - 1 = (1/8)(y-2)^2


So the final equation is now in the form x-a=(b/c)(y-d)^e where a = 1, b = 1, c = 8, d = 2 and e = 2

Answer 4

Thank you sooo much!! =) So my equation should be x-1=(1/8)(y-2)^2
Which would then become x=(1/8)(y-2)+1^2 correct?

Answer 5

yw, if you meant x=(1/8)(y-2)^2+1, then yes, that's another way of writing it