Question

What is the equation of the parabola that has a vertex at `(-4, 16)` and passes through point `(0,0)`?

Answer 1

Let us solve this problem by substituing both points into a parabola equation: `ax^2 + b x + c = y(x)`

Explanation:

• First of all, let us substitute `(0,0)`:

`ax^2+bx+c= y(x) rightarrow a cdot 0^2 + b cdot 0 + c = y(0) rightarrow c = 0`

Thus, we obtain the independent term in equation, getting `ax^2 + bx = y(x)`.

• Now, let us substitute the vertex, `(-4, 16)`. We get:

`a cdot (-4)^2 + b cdot (-4) = 16 rightarrow 16 a - 4 b = 16 rightarrow 4 a - b = 4`

Now, we have a relation between `a` and `b`, but we cannot determine them uniquely. We need a third condition.

• For any parabola, the vertex can be obtained by:

`x_"vertex" = {-b}/{2a}`

In our case:

`x_"vertex" = -4 = {-b}/{2a} rightarrow b = 8 a`

• Finally, we must solve the system given by:

`{4a-b=4 ; b = 8a}`

Replacing `b` from second equation to the first one:

`4a-(8a)=4 rightarrow -4 a = 4 rightarrow a = -1`

And finally:

`b = -8`

This way, the parabola equation is:

`y(x) = -x^2 - 8x`