Forms of a Quadratic Relation

Standard form is what you get after expanding and collecting like terms. Setting $x=0$ wipes out the $a{x}^2$ and $bx$ terms and leaves $y=c$, so $c$ is the $y$-intercept. The leading coefficient $a$ controls the shape: $a>0$ opens up (the vertex is a minimum), $a<0$ opens down (the vertex is a maximum), and a larger $|a|$ makes the parabola narrower because the $y$-values grow faster as you move away from the vertex. The axis of symmetry is hidden, but a neat argument finds it. Factor $x$ out of the first two terms: $y=x(ax+b)+c$. This equals $c$ whenever $x=0$ or $x=-\frac{b}{a}$, so those two points are at the same height. A parabola is symmetric, so its axis sits exactly halfway between them: $x=\frac{0+(-b/a)}{2}=-\frac{b}{2a}$. Substituting that $x$ back in gives the vertex. For $y=2x^2-12x+7$: $x=-\frac{-12}{2(2)}=3$ and $y=2(9)-36+7=-11$, so the vertex is $(3,-11)$.

Factored form works because of the zero product property: if a product equals zero, at least one of its factors must be zero. Setting $y=0$ in $y=a(x-r_1)(x-r_2)$ forces $x-r_1=0$ or $x-r_2=0$, so the zeros (roots, $x$-intercepts) are $x=r_1$ and $x=r_2$. Watch the signs: the bracket $(x+4)$ gives the zero $x=-4$, and a bracket like $(2x-3)$ gives $x=\frac{3}{2}$, not $3$. The axis of symmetry is the midpoint of the zeros, $x=\frac{r_1+r_2}{2}$, and substituting it gives the vertex. For $y=2(x-1)(x-5)$: the axis is $x=\frac{1+5}{2}=3$, and $y=2(3-1)(3-5)=2(2)(-2)=-8\text{,}$ so the vertex is $(3,-8)$. The $y$-intercept is quick too: $x=0$ gives $2(-1)(-5)=10$. Not every parabola has a factored form using real numbers. A parabola that never touches the $x$-axis has no real zeros, so there are no real brackets to write; you will learn to detect this with the discriminant.

Interpreting a quadratic in factored form

Vertex form is the parent function $y=x^2$ after transformations: a vertical stretch by $a$ (a reflection too if $a<0$), a horizontal shift of $h$ and a vertical shift of $k$. That is why the vertex moves from $(0,0)$ to $(h,k)$. There is also an algebraic reason. A square is never negative, so ${(x-h)}^2\ge 0$, with equality only when $x=h$. If $a>0$, the term $a{(x-h)}^2$ is smallest (zero) at $x=h$, so the smallest possible $y$ is $k$: a minimum value of $k$. If $a<0$, the term is at most zero, so the largest $y$ is $k$: a maximum value. Read $h$ carefully: $y=-2{(x+3)}^2-7$ is $y=-2{(x-(-3))}^2+(-7)$, so the vertex is $(-3,-7)$. Vertex form also tells you the number of zeros without solving: if $a$ and $k$ have the same sign, there are no zeros (for $a>0$ and $k>0$ the lowest point is above the $x$-axis; for $a<0$ and $k<0$ the highest point is below it); if $k=0$, the vertex sits on the axis and there is exactly one; if $a$ and $k$ have opposite signs, there are two.

Introduction to vertex form of a quadratic

• Factored or vertex → standard: expand. Multiply the brackets first, then distribute $a$, then add $k$ last. Collect like terms.
• Standard → factored: take out any common factor, then factor the trinomial. Always check by expanding back.
• Standard → vertex: complete the square (next lesson), or use the shortcut: find $h=-\frac{b}{2a}$, substitute to get $k$, and write $y=a{(x-h)}^2+k$ with the same $a$.
• Factored → vertex: $h$ is the midpoint of the zeros, $k$ is the $y$-value there, and $a$ is unchanged. For $y=2(x-1)(x-5)$ that gives $y=2{(x-3)}^2-8$.