How can I calculate properly the graph of a quadratic function?

Question

I want to know how to calculate this: (I don't know his name in english, or in spanish, our teacher call it "the little belly" :)) We've just today had the math final in school and we had to draw a function very similar like that one, our teacher say that we must write that part as we wish because it's too complicate to for us, she say that maybe at the end of the next year we would teach how to calculate this, but I don't have enought patience...

Actually I know this:
$$x^{2}+x-2$$

$\text{Domain} = \mathbb{R} \text{ or } (-\infty;+\infty)$
$\text{Roots (Intersection to x)} = (1;0) \text{ and }(-2;0)$ $$\frac{-1±\sqrt{1^{2}-4\times 1 \times (-2)}}{2\times 1}=0$$ $$\frac{-1±\sqrt{1+8}}{2}=0$$ $$\frac{-1±\sqrt{9}}{2}=0$$ $$\frac{-1±3}{2}=0$$ $$x_{1}=\frac{-1+3}{2}=\frac{2}{2}=1$$ $$x_{2}=\frac{-1-3}{2}=\frac{-4}{2}=-2$$ $\text{Sorted to Source (Intersection to y)} = (0;-2)$ $$0^{2}+0-2=0+0-2=-2$$ $\text{Vertex} = (-\frac{1}{2};-\frac{9}{4})$ $$x_{v}=\frac{-1}{2\times1}=\frac{-1}{2}=-\frac{1}{2}$$ $$y_{v}=(-\frac{1}{2})^{2}+(-\frac{1}{2})-2$$ $$y_{v}=(-\frac{1}{4})-\frac{1}{2}-2$$ $$y_{v}=(-\frac{1}{4})-\frac{1}{2}-2$$ $$y_{v}=-\frac{9}{4}$$

And finally (In spanish we use the letter $C$, but I am not sure if in english it would be $S$ (set instead of "conjunto")):

• $\text{Image} = (+\infty;-\frac{9}{4})$
• $C^{+} = (-\infty;-2) U (1;+\infty)$
• $C^{-} = (-2;1)$
• $C^{↑} = (-\frac{1}{2};+\infty)$
• $C^{↓} = (-\infty;-\frac{1}{2})$
• $C^{0} = \text{{-2;1}}$

But with all this values I only know:

• Where the function pass throught the y.
• Where the function pass throught the x (two times).
• And where the function stop decreasing and start increasing (also the bottom of it, where it's the "belly").

But with that information I can only make straight lines, not curves like in the picture. How can I know how to make them curves?

With my current information I can only do (in red):

My only idea is to use a table of values, but even with that it can't be very precise, or at least without an huge effort (using a table with a lot of values).

Answer 1

Rounding our evaluation to the nearest half grid resolution ($0.25$), we trivially realize that:

• The center of the parabola is $(x_0,y_0)=(-0.5,-2.25)$,
• The parabola passes through $(2,4)$
• The parabola passes through $(-3,4)$, hence it is a y-axis parabola.

From the equation of the parabola: $$ y-y_0=\frac{1}{4p}(x-x_0)^2 $$ we evaluate and find the focus: $$ 4+2.25=\frac{1}{4p}(2+0.5)^{2}\\ 6.25 * 4p=1*(2+0.5)^{2}\\ 25p = 6.25\\ p = \frac{6.25}{25}= \frac{1}{4} $$ Hence the equation is: $$ y=(x+0.5)^2-2.25 $$

From here you are ready. You can simply have a table for each parabola point in the given grid. Note the symmetry on the y values, confirming the parabola has the vertical line as axis: $$ \begin{array}{c} x& y\\ -3& 4\\ -2.5& 1.75\\ -2& 0\\ -1.5& -1.25\\ -1& -2\\ -0.5& -2.25\\ 0& -2\\ 0.5& -1.25\\ 1& 0\\ 1.5& 1.75\\ 2& 4 \end{array} $$

Or draw it in any mathematical package like Wolfram Alpha.

If you indeed want to draw it by hand, the parabola focus is: $(x_0,y_0+p)=(-0.5,-2)$ and the directrix is $y=y_0-p=-2.5$, and then the artwork follows as shown here.