The product form and the standard form

A quadratic function can be written in three forms, and each one shows other properties at once:

• the vertex form $\displaystyle {y}={a(x-b)^{2}+k}$ – the vertex;
• the product form $\displaystyle {y}={a(x-x_1)(x-x_2)}$ – the intersection points with the $\displaystyle x$-axis;
• the standard form $\displaystyle {y}={ax^{2}+bx+c}$ – the intersection point with the $\displaystyle y$-axis.

In $\displaystyle {y}={a(x-x_1)(x-x_2)}$:

• the intersection points with the $\displaystyle x$-axis are at $\displaystyle x_1$ and $\displaystyle x_2$: a product is zero when one of its factors is zero;
• the vertex is on the axis of symmetry, halfway between them: its $\displaystyle x$ is $\displaystyle \frac{x_1+x_2}{2}$.

For example, $\displaystyle {y}={(x-1)(x-5)}$:

• the $\displaystyle x$-axis: $\displaystyle A(1,0)$, $\displaystyle B(5,0)$;
• the vertex: $\displaystyle {x}={\frac{1+5}{2}}={3}$, $\displaystyle {y}={(3-1)(3-5)}={-4}$ – $\displaystyle V(3,-4)$, a minimum point;
• the $\displaystyle y$-axis: $\displaystyle {y}={(0-1)(0-5)}={5}$ – $\displaystyle C(0,5)$.

$\displaystyle {y}={-(x+3)(x-1)}$: here $\displaystyle {a}={-1}$, so the vertex is a maximum point.

• The $\displaystyle x$-axis: $\displaystyle A(-3,0)$, $\displaystyle B(1,0)$.
• The vertex: $\displaystyle {x}={\frac{-3+1}{2}}={-1}$, $\displaystyle {y}={-(-1+3)(-1-1)}={4}$ – $\displaystyle V(-1,4)$.
• The $\displaystyle y$-axis: $\displaystyle {y}={-(0+3)(0-1)}={3}$ – $\displaystyle C(0,3)$.

In $\displaystyle {y}={ax^{2}+bx+c}$ ($\displaystyle a\neq0$):

• the intersection point with the $\displaystyle y$-axis is $\displaystyle (0,c)$: at $\displaystyle {x}={0}$ only $\displaystyle c$ is left;

• the $\displaystyle x$ of the vertex:

  $\displaystyle x$ $\displaystyle =$ $\displaystyle -\frac{b}{2a}$$\displaystyle a\neq0$ ⇄

The vertex comes from the product form. Opening its brackets gives $\displaystyle {a(x-x_1)(x-x_2)}={ax^{2}-a(x_1+x_2)x+ax_1x_2}$, so $\displaystyle {b}={-a(x_1+x_2)}$, and the vertex’s $\displaystyle x$, $\displaystyle \frac{x_1+x_2}{2}$, is $\displaystyle -\frac{b}{2a}$. The rule holds for every parabola, even one that does not meet the $\displaystyle x$-axis.

The extended distributive law opens the product form: $\displaystyle {(x-1)(x-5)}={x^{2}-6x+5}$. In the standard form the same parabola shows $\displaystyle {c}={5}$ – the point $\displaystyle (0,5)$ – and the vertex’s $\displaystyle x$, $\displaystyle {-\frac{-6}{2\cdot1}}={3}$, as found before.

$\displaystyle {y}={x^{2}-4x+1}$, in four steps:

• $\displaystyle {a}={1>0}$: the vertex is a minimum point.
• The vertex: $\displaystyle {x}={-\frac{-4}{2\cdot1}}={2}$, $\displaystyle {y}={2^{2}-4\cdot2+1}={-3}$ – $\displaystyle V(2,-3)$.
• The $\displaystyle y$-axis: $\displaystyle C(0,1)$.
• The point symmetric to $\displaystyle C$ across the axis $\displaystyle {x}={2}$: $\displaystyle D(4,1)$.

Its intersection points with the $\displaystyle x$-axis are not whole numbers, so factoring with whole numbers cannot find them.

For each parabola find the intersection points with the axes and the vertex.
Then open the brackets to get the standard form.

$\displaystyle {y}={(x-2)(x-6)}$

$\displaystyle {y}={-(x-1)(x+3)}$

$\displaystyle {y}={2x(x-4)}$

For each parabola find the vertex and the intersection point with the $\displaystyle y$-axis.
Factor its expression, and find the intersection points with the $\displaystyle x$-axis.

$\displaystyle {y}={x^{2}-6x+8}$

$\displaystyle {y}={-x^{2}+4x+5}$

$\displaystyle {y}={2x^{2}-12x+10}$

The parabola is $\displaystyle {y}={x^{2}+2x+4}$.
Find its vertex.
Does it meet the $\displaystyle x$-axis?
Explain.

A parabola meets the $\displaystyle x$-axis at $\displaystyle (-2,0)$ and $\displaystyle (4,0)$, and the $\displaystyle y$-axis at $\displaystyle (0,-8)$.
Write its equation in the product form, and open it to the standard form.