The equation $y = x^2 + \frac{1}{2}ax + a$ represents a parabola for all real values of $a$

Question

The task is to show that any parabola with the value a passes through shared point.

I graphed the parabola and understand that the shared point is $(-2,4)$. My current issue is generalizing why this is and how to prove it.

Answer 1

Just a disclaimer this is not the most formal proof so take it with some skepticism

Considered rearranging the quadratic so that $a$ is the subject. You get $a=\frac{y-x^2}{x/2+1}$ since $a$ can take any real value then the only candidates for $x$ and $y$ must make the right hand side an indeterminate form, e.g. since $x$ and $y$ are both finite both neumerator and denominator must be zero. So we want $x/2+1=0$ which gives $x=-2$ and $y-x^2=0$ which gives $y=4$. All that's left is to substitute these values into the original equation to verify that this is indeed a shared point.

Answer 2

Hint: If $x=-2$, what is the value of $y$ if you know that the point $(x,y)$ lies on the parabola?

Answer 3

Let $a,b \in \mathbb R$ and $a \ne b$. Schow that the equation

$$ x^2 + \frac{1}{2}ax + a= x^2 + \frac{1}{2}bx + b$$

has the unique solution $x=-2.$

Answer 4

Just a note that $$x^2+\frac{1}{2}ax+(a-4)\\=x^2+2x-2x+\frac{1}{2}ax+a-4\\=x(x+2)-\left(2-\frac{1}{2}a\right)(x+2)\\=(x+2)\left(x-2+\frac{1}{2}a\right)$$

And hence, you get 2 points which equal 4 for the original function: $x=-2$ and $x=2-\frac{1}{2}a$.

In general, there is such common points when you have some expression $x^2+f(a) x+g(a)$.