Quadratic Functions

A quadratic function is a function of the form

$$f(x) = ax^2 + bx + c.$$

I've already discussed quadratic functions a little; you know that you can use the graph of a quadratic function is a parabola. The parabola opens upward if $a
   > 0$ and opens downward if $a < 0$ . Knowing which way the parabola opens and where the roots are gives a pretty good picture of the graph.

The vertex of a parabola is the "tip" of the graph --- the lowest point on a parabola that opens up and the highest point on a parabola that opens down.

$$\hbox{\epsfysize=2in \epsffile{quadratic-functions1.eps}}$$

$$\hbox{\epsfysize=1.75in \epsffile{quadratic-functions2.eps}}$$

The formula $x =
   -\dfrac{b}{2a}$ is easy to remember, since it's the first "piece" of the quadratic formula:

$$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$


Example. Find the coordinates of the vertex of $y = (x - 2)(x + 7)$ .

The roots are $x = 2$ and $x = -7$ . The vertex is halfway between the roots, which is at

$$x = \dfrac{2 + (-7)}{2} = -2.5.$$

The y-coordinate of the vertex is

$$y = (-2.5 - 2)(-2.5 + 7) = -20.25.$$

Therefore, the vertex is $(-2.5, -20.25)$ .


Example. Find the coordinates of the vertex of $y = 2x^2 + 5x + 17$ .

Since it would be too hard to find the roots by factoring, I'll use the formula $x =
   -\dfrac{b}{2a}$ instead. $b = 5$ and $a =
   2$ , so the x-coordinate of the vertex is

$$x = -\dfrac{5}{2\cdot 2} = -1.25.$$

The y-coordinate of the vertex is

$$y = 2(-1.25)^2 + 5(-1.25) + 17 = 13.875.$$

The vertex is $(-1.25,
   13.875)$ .


If the vertex of a parabola is $(h,k)$ , the parabola's equation has the form

$$y = a(x - h)^2 + k.$$

This checks, since plugging $x = h$ in gives

$$y = a(h - h)^2 + k = 0 + k = k.$$

$y = a(x - h)^2 + k$ is called the standard or vertex form of the equation of a parabola.


Example. Find the standard/vertex equation of the parabola whose equation is $y = x^2 + 6x + 10$ . What are the coordinates of the vertex?

The idea is to complete the square in x. $\dfrac{6}{2} = 3$ , and $3^2 = 9$ . Therefore, I write the equation as

$$y = x^2 + 6x + 10 = x^2 + 6x + 9 + 1 = (x + 3)^2 + 1.$$

The vertex is $(-3,-1)$ .


Example. The vertex of a parabola is $(3,1)$ . The parabola passes through the point $(4,3)$ . Find the equation of the parabola.

Since the vertex is $(3,1)$ , the equation is

$$y = a(x - 3)^2 + 1.$$

Plug in $x = 4$ , $y = 3$ :

$$3 = a(4 - 3)^2 + 1, \quad 3 = a + 1, \quad a = 2.$$

The equation is $y = 2(x -
   3)^2 + 1$ .


Example. Find the dimensions of the rectangle with the largest area that has a perimeter of 20 feet.

$$\hbox{\epsfysize=1.75in \epsffile{quadratic-functions3.eps}}$$

Let x be the width of the rectangle and let y be the height. The area is

$$A = xy.$$

The perimeter is 20, so

$$20 = 2x + 2y.$$

Then

$$\matrix{& 20 & & & = & 2x & + & 2y \cr / & 2 & & & & 2 & & 2 \cr \noalign{\vskip2pt\hrule\vskip2pt} & 10 & & & = & x & + & y \cr - & & & y & & & & y \cr \noalign{\vskip2pt\hrule\vskip2pt} & 10 & - & y & = & x & & \cr}$$

Substitute $x = 10 - y$ into $A = xy$ :

$$A = (10 - y)y = 10y - y^2.$$

The graph of A is a parabola opening downward. The roots are $y = 0$ and $y =
   10$ . The vertex is at $y = 5$ , and it's the highest point on the graph. Therefore, this value gives the largest value of A. When $y = 5$ , $x = 10 -
   5 = 5$ .

Thus, the rectangle of largest area is 5 feet wide and 5 feet tall --- a square.


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