finding the zeros of a quadratic function

Question

I am learning quadratic equations and got stuck at an exercise which asks to find the minimum point of this parabola:

The solution starts off by stating that:

The parabola has zeros at $x = -2.2$ and $x = 0.6$

∴ $f(x) = a(x + 2.2)(x - 0.6) = a(x^2 + 1.6 x - 1.32)$

I think, I am familiar with the product rule and basic rules of transforming a function (such as shifting to left or right), but cannot understand where the a comes from into the equation. Can anyone please explain how you know that there should be some number in front of the parentheses?

The exercise is from Math is Fun: Quadratic Equation Graphing, Question 9 (scroll down)

Answer 1

If the graph of function $f$ is a parabola then its universal form is:$$f(x)=ax^2+bx+c$$ where $a,b,c$ denote constants and $a\neq 0$.

If it is known that it has zeros $x_1$ and $x_2$ exist then we can write this also as:$$f(x)=a(x-x_1)(x-x_2)$$where $a,x_1,x_2$ are constants and $a\neq 0$.

If you receive as further info that it has zeros at $x=-2.2$ and $x=0.6$ then you arrive at: $$f(x)=a(x+2.2)(x-0.6)$$where $a$ is a constant and $a\neq 0$.

Answer 2

Because the general polynomial representation of $n$ degree polynomial is

$$f(x)=a (x-x_0)(x-x_1)(x-x_2)....(x-x_{n-1}).$$ Here, $a\ne 0$ and $a$ is the coefficient of $x^n.$

---

Also in your question, suppose you haven't multiplied by $a$; then check the value of the polynomial at $x=0$. Then your polynomial gives $-1.32$.

But see the value in the graph; it is $-33$ which means your polynomial is wrong.

Now consider the case when $a$ is multiplied in front of the polynomial. You get $a\cdot(-1.32)=-33.$ Thus you get your value of $a$.

Answer 3

Because the general equation for the parabola in canonical form is

$$y=ax^2+bx+c$$

and all the parabolas with equation

$$y=a(x+2.2)(x-0.6)$$

have the same zeros at $x=-2.2$ and $x=0.6$.

Answer 4

The general form of a quadratic function is $$ y = a x^2 + b x + c $$ If you have sample points $(x_i, y_i)$ from the graph you get equations $$ x_i^2 a + x_i b + 1 \cdot c = y_i $$ which are linear equations in the unknowns $a, b, c$.

You need at least three different of them to be able to solve for the unknown parameters. Which means you need to have three different sample points.

This is probably the reason why your image has three points given, the two roots and the point on the $y$-axis.

The form you work with is derived from an algebraic result, which in the case of a quadratic function is $$ y = a (x - x_1) (x - x_2) $$ where $a$ is a scaling factor and $x_1$ and $x_2$ are roots of the function, thus having the property $y(x_1) = y(x_2) = 0$.

If you use the two points for the roots you end up with $$ 0 = a (x_1-x_1) (x_1-x_2) \\ 0 = a (x_2-x_1) (x_2-x_1) $$ which are true for any $a$, so you need a third point $(x_3, y_3)$ from the graph to determine $a$. $$ y_3 = a (x_3-x_1)(x_3-x_2) $$ It should be different from the other two, to provide you $a$. Then $$ a = \frac{y_3}{(x_3-x_1)(x_3-x_2)} $$

Answer 5

As the parabola is symmetric, the minimum is halfway between the roots, at $x=-0.8$, with value

$$a((-0.8)^2 + 1.6 (-0.8) - 1.32)=-1.96a.$$

Answer 6

The coefficient $a$ is there because, when you know two distinct roots , say α and β of a polynomial $f(x)$, you know this polynomial is divisible by $x-α$ and by $x-β$, hence by their product $(x-α)(x-β)$ since these factors are coprime. Thus one may write $$f(x)=g(x)(x-α)(x-β),\qquad\deg g=\deg f-2.$$

This means here that $g(x)$ has degree $0$, i.e. $g(x)$ is a constant. This constant can be determined from the value of $f(0)$.

Last point: the extremum of a quadratic function which has roots is obtained at the arithmetic mean of the roots.