Parabolas: reading a quadratic relationship

You have 20 m of fencing for a rectangular plot. Do sides of 3 and 7 m give the same area as 5 and 5 m? The perimeter is equal, but the areas are 21 and 25 m². A quadratic relationship shows where enlarging one dimension stops helping the area.

Use quadratic equations and functions. Fencing covers all four sides; thickness and openings are ignored.

A vertex from a nonnegative square

Let $x$ be numerical side length in meters. From $2(x+y)=20$ , the other side is $y=10-x$ . Both must be positive, so $0<x<10$ . Numerical area in m² is

S(x)=x(10-x)=25-(x-5)^2.

A square is nonnegative, so $S(x)\le25$ , with equality only at $x=5$ . The plot is a 5×5 m square, and $(5,25)$ is the area graph’s vertex. Inputs $x=5-d$ and $x=5+d$ give the same $d^2$ and area: the graph is symmetric about $x=5$ .

A quadratic function’s graph is a parabola. The negative coefficient of the square here makes it open downward, so its vertex is a maximum. Zero area at 0 and 10 represents degenerate rectangles; those endpoints are outside the model. The area range is $(0,25]$ .

How coefficients change the shape

The form $f(x)=a(x-h)^2+k$ , $a\ne0$ , is convenient to read. Its vertex is $(h,k)$ and symmetry axis $x=h$ . For $a>0$ the vertex is a minimum; for $a<0$ , a maximum. At a given horizontal offset from the vertex, vertical displacement equals $a$ times that offset squared.

To turn $x^2$ into $2(x-1)^2-3$ , shift right by 1, double vertical distances from the zero level, then shift down by 3. The vertex becomes $(1,-3)$ . Multiplying the entire resulting expression by −1 would reflect it across the horizontal axis. Quadratic forms, OpenStax §5.1.

Shift and stretch

Area with a 20 m perimeter

Top: solid y=x², dashed y=2(x−1)²−3, vertex (1,−3), abstract units. Bottom: S=25−(x−5)², maximum (5,25), open endpoints 0 and 10 excluded. Uniform scales; portions of curves, not physical trajectories.

The vertex in ax²+bx+c form

Completing the square for $a\ne0$ gives

ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.

Hence $h=-b/(2a)$ and $k=f(h)=c-b^2/(4a)$ . Keep the minus sign in the horizontal coordinate. For $S=-x^2+10x$ , $a=-1$ , $b=10$ , $c=0$ , so $h=5$ , $k=25$ .

These coordinates describe the full parabola’s vertex. If the problem allowed only $1\le x\le3$ , the vertex at 5 would be inaccessible; the largest area on that domain would be 21 m² at $x=3$ .

Zeros answer equations; signs answer inequalities

Zeros are inputs with $f(x)=0$ , found by a quadratic equation. Two distinct real roots give two horizontal-axis crossings; a double root gives tangency; a negative discriminant gives no real zeros.

For $f(x)=2(x-1)^2-3$ , the condition $f(x)\le0$ is equivalent to $(x-1)^2\le3/2$ . Thus

1-\sqrt{3/2}\le x\le1+\sqrt{3/2}.

This is the part of the parabola on or below the axis. There is no need to guess signs from a curve’s appearance: roots locate the boundaries, while opening direction or factoring confirms intervals.

Perimeter became a formula, the formula became a square, and the square gave an exact area bound. Reading a vertex helps with other quadratic models too, provided their domains and units are checked first.