# Conceptual question on equations of the form ##x=ay^2+by+c##

• chwala
In summary, quadratic equations are equations with the form ##y=ax^2+bx+c## where ##a,b##, and ##c## are constants and ##x,y## are variables. The equation ##x=y^2+2y+1## can also be viewed as a quadratic equation by switching the variables, but in this case, ##x## is a function of ##y## while ##y## is not a function of ##x##. This means that for any value of ##x##, we can obtain one unique value for ##y##, but the same is not true for any value of ##y##.
chwala
Gold Member
Homework Statement
This is my own question;
Relevant Equations
Now i just need some clarification; we know that quadratic equations are equations of the form ##y=ax^2+bx+c## with ##a,b## and ##c## being constants and ##x## and ##y## variables.

Now my question is... can we also view/look at ##x=y^2+2y+1## as quadratic equations having switched the variables ? thanks...

chwala said:
Homework Statement: This is my own question;

Now i just need some clarification; we know that quadratic equations are equations of the form ##y=ax^2+bx+c## with ##a,b## and ##c## being constants and ##x## and ##y## variables.

Now my question is... can we also view/look at ##x=y^2+2y+1## as quadratic equations having switched the variables ? thanks...
Yes. Of course we can.

However, I hope you realize that in this case, ##x## is a function of ##y##, but ##y## is not a function of ##x##.

MatinSAR and chwala
SammyS said:
Yes. Of course we can.

However, I hope you realize that in this case, ##x## is a function of ##y##, but ##y## is not a function of ##x##.
yes @SammyS ...we now have ##x## as the dependent variable......but is the relation going to be a Function? as we require to obtain one unique value( one and only one) for ##y## for any value of ##xε\mathbb{R}##.

Last edited:
it is a function of variable ##y##

chwala and malawi_glenn
chwala said:
yes @SammyS ...we now have ##x## as the dependent variable......but is the relation going to be a Function? as we require to obtain one unique value( one and only one) for ##y## for any value of ##xε\mathbb{R}##.
I believe that I answered this in Post #2 .

chwala and Mark44
chwala said:
as we require to obtain one unique value( one and only one) for ##y## for any value of ##xε\mathbb{R}##.
But this is needed when ##y## is a function of ##x##.
If we have ##x=y^2## we can say that ##x## is a function of ##y##. But ##y## is not a function of ##x## because: ##y=\pm x^\frac {1}{2}##.

chwala

## What is the general form of the equation ##x = ay^2 + by + c##?

The general form of the equation ##x = ay^2 + by + c## is a quadratic equation in terms of ##y##, where ##a##, ##b##, and ##c## are constants. The variable ##x## is expressed as a quadratic function of ##y##.

## How do you determine the vertex of the parabola represented by ##x = ay^2 + by + c##?

The vertex of the parabola represented by ##x = ay^2 + by + c## can be found using the formula for the vertex of a quadratic equation. The ##y##-coordinate of the vertex is given by ##y = -\frac{b}{2a}##. Once you have the ##y##-coordinate, substitute it back into the equation to find the corresponding ##x##-coordinate.

## What is the axis of symmetry for the equation ##x = ay^2 + by + c##?

The axis of symmetry for the equation ##x = ay^2 + by + c## is a vertical line that passes through the vertex of the parabola. The equation for the axis of symmetry is ##y = -\frac{b}{2a}##.

## How can you determine if the parabola opens to the left or to the right?

The direction in which the parabola opens is determined by the sign of the coefficient ##a##. If ##a > 0##, the parabola opens to the right. If ##a < 0##, the parabola opens to the left.

## How do you find the roots of the equation ##x = ay^2 + by + c##?

To find the roots of the equation ##x = ay^2 + by + c##, set ##x## equal to zero and solve the resulting quadratic equation ##ay^2 + by + c = 0## for ##y##. You can use the quadratic formula ##y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}## to find the values of ##y## that satisfy the equation.

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