Characteristics of a quadratic curve

Characteristics of a quadratic curve #

In this lesson we will be exploring four different characteristics of a quadratic curve.

  1. Concavity
  2. y-Intercept
  3. Vertex (turning point)
  4. Equation of the line of symmetry

Characteristics of a quadratic curve

Concavity #

Before delving into the concavity of a quadratic curve, lets investigate what is concavity of a function over an interval. Simply put, the rate at which the slope of a function changes is called its concavity over an interval.

Smelling second derivative, aren’t you!

The slope of a function at a given point is obtained by its first derivative and the rate of change of slope is simply its second derivative.

Lets put it in the context of a quadratic function:

f(x)=ax2+bx+cf(x) = ax^2 + bx + c

The first and second derivatives:

f′(x)=2ax+bf′′(x)=2a\begin{aligned} f'(x) &= 2ax + b \\ f''(x) &= 2a \\ \end{aligned}

The second derivative only depends on the value of a a .

If a a is positive, a>0 a > 0 , then the curve opens up and it is called concave up and if a a is negative, a<0 a < 0 , the curve opens down and is called concave down.

y-Intercept #

The point where a curve intersects the y y -axis is called y-intercept. By setting the value of x x to 0 0 we can find the y-intercept.

For quadratic function:

f(x)=ax2+bx+cf(0)=a(0)2+b(0)+cf(0)=c\begin{aligned} f(x) &= ax^2 + bx + c \\ f(0) &= a(0)^2 + b(0) + c \\ f(0) &= c \end{aligned}

the value of c c gives the y-intercept. The point where the quadratic curve cuts the y y -axis is (0,c) (0, c) .

Vertex #

The turning point of a curve. If a quadratic function is concave up, it will have a minimum point and if it is concave down, it will have a maximum point.

The maximum or a minimum points can be obtained by taking the first derivative of a function and setting it to 0 0 .

First derivative of a quadratic function is:

f(x)=ax2+bx+cf′(x)=2ax+b\begin{aligned} f(x) &= ax^2 + bx + c \\ f'(x) &= 2ax + b \end{aligned}

Setting it to 0 0 gives the value of x x ,

f′(x)=02ax+b=02ax=−bx=−b2a\begin{aligned} f'(x) &= 0 \\ 2ax + b &= 0 \\ 2ax &= -b \\ x &= -\frac{b}{2a} \end{aligned}

Solving for f(−b2a) f(-\frac{b}{2a}) to get the y y -coordinate of the turning point,

f(−b2a)=a(−b2a)2+b(−b2a)+cf(−b2a)=a(b24a2)−(b22a)+cf(−b2a)=b24a−b22a+c\begin{aligned} f(-\frac{b}{2a}) &= a(-\frac{b}{2a})^2 + b(-\frac{b}{2a}) + c \\ f(-\frac{b}{2a}) &= a(\frac{b^2}{4a^2}) - (\frac{b^2}{2a}) + c \\ f(-\frac{b}{2a}) &= \frac{b^2}{4a} - \frac{b^2}{2a} + c \end{aligned}

Make the denominator of every fraction equal to 4a 4a to make life easy,

f(−b2a)=b24a−2b24a+4ac4af(−b2a)=b2−2b2+4ac4af(−b2a)=−b2+4ac4af(−b2a)=−b24a+c\begin{aligned} f(-\frac{b}{2a}) &= \frac{b^2}{4a} - \frac{2b^2}{4a} + \frac{4ac}{4a} \\ f(-\frac{b}{2a}) &= \frac{b^2 - 2b^2 + 4ac}{4a} \\ f(-\frac{b}{2a}) &= \frac{-b^2 + 4ac}{4a} \\ f(-\frac{b}{2a}) &= -\frac{b^2}{4a} + c \end{aligned}

So the turning point, vertex, of a quadratic curve is given by:

(−b2a,−b24a+c) (-\frac{b}{2a}, -\frac{b^2}{4a} + c)

Equation of the Line of Symmetry #

If you fold a curve along an axis such that one half is identical to the other half then the axis is called the line of symmetry.

A quadratic function always has a line of symmetry parallel to the y y -axis and it passes through the turning point.

So the equation of the line of symmetry for a quadratic curve is given by:

x=−b2a x = -\frac{b}{2a}