Plotting a Parabola using a pencil, graph book and a calculator

Question

How do I go about plotting the equation: $$y^2 + 4x - 2y + 3 = 0$$

I am in my first year in the university. I am taking a course on Analytical Geometry and formulas are all over the place. I could just cram the formulas for questions, but there are so many, plus I prefer to understand what I am studying and keeping it in my long term memory.

I pretty much understand most calculations about a line, but something seems complicated about conical sections. I am trying to sketch this parabola by hand to see if I can get a better feel for what a parabola is.

$$y^2 - 2y = -4x - 3$$ $$y(y-2) = -4x - 3$$

I would take a similar step if I wanted to sketch a line, but there is no way (or at least not one obvious to me) to express y as an explicit function of x so that I can create a table of values from which I can plot my graph.

Answer 1

HINT…if you are familiar with the standard form of the parabola $$y^2=4ax$$ and you rewrite your equation as $$(y-1)^2=-4\left(x+\frac12\right)$$ Then you can consider your graph as a sequence of linear transformations of the standard form.

Answer 2

The standard form for a parabola is

$$y = a(x-h)^2 + k, \tag{1}$$ where $a$ is a nonzero scaling constant, and $(h,k)$ is the coordinate of the parabola's vertex. In your case, however, the role of the $x$ and $y$ coordinates are switched, so your parabola has the form

$$x = a(y-k)^2 + h, \tag{2}$$ where again $(h,k)$ is the vertex. Notice that when I switched $x$ and $y$, I also switched $h$ and $k$; otherwise, the vertex would become $(k,h)$. Now when the $x$ and $y$ axes are switched, the parabola does not open up or down, but rather, to the left (for $a < 0)$ or to the right (for $a > 0$).

I leave it as an exercise for you to determine the corresponding values of $a, h, k$ for your particular parabola and to write it in standard form $(2)$.

As for the matter of sketching such a parabola, you would first draw the vertex $(h,k)$. Then, because your parabola opens sideways, draw a horizontal line through the vertex, parallel to the $x$ axis. This will be the axis of symmetry for the parabola.

Next, note that when $y = k \pm 1$, we have $x = a + h$. So you would plot the points $(a+h, k+1)$ and $(a+h, k-1)$ as belonging to the parabola. Similarly, when $y = k \pm 2$, we have $x = 4a + h$, and you would plot $(4a+h, k+2)$ and $(4a+h, k-2)$. This now gives you five points on the parabola, through which you should be able to sketch the desired curve.

Answer 3

plotting points: which is what they asked

$$ x = - \left( \frac{(y-1)^2 +2}{4} \right) $$