Can you measure height with a stopwatch? How could that be possible?

This lesson is all about quadratic functions, like the one used in Physics and described later in the lesson, which allows a good approximation of the height of a tall object to be calculated with time!

### Methods for Solving Quadratic Functions

A quadratic function can be described as:

Formally: A function \begin{align*}f\end{align*} defined by \begin{align*}f(x)=ax^{2}+bx+c\end{align*}, where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} are real numbers and \begin{align*}a \ne 0\end{align*}.

Informally: The defining characteristic of a quadratic function is that it is a polynomial whose highest exponent is 2.

There are several ways to write quadratic functions:

**standard form**, the form of the quadratic function above: \begin{align*}f(x)=ax^{2}+bx+c\end{align*}**vertex form**, commonly used for quick sketching: \begin{align*}f(x)=a(x-h)^{2}+k\end{align*}**factored form**, excellent for finding*x*-intercepts: \begin{align*}f(x)=a(x-r_{1})(x-r_{2})\end{align*}

You can convert between forms of quadratic functions using algebra and you will see that there are uses for each of these forms when working with quadratic functions.

When graphing, the **\begin{align*}y-\end{align*}**** intercept** of a quadratic function in standard form is \begin{align*}(0, c)\end{align*} and it is found by substituting for \begin{align*}x\end{align*} in \begin{align*}f(x)=ax^{2}+bx+c\end{align*}.

### Examples

#### Example 1

Earlier, you were asked if you could measure height with a stopwatch.

What formula could be used to calculate the height of an object based on the time required for an object to fall from the top?

- One formula is \begin{align*}h = \frac{1}{2}gt^{2}\end{align*}

This is not completely accurate, as it ignores the effects of air resistance of an object, but offers a very close approximate for most common objects.

The value *g* is gravity. In Metric units this is \begin{align*}9.81 m/s/s\end{align*}

The value *t* is the time taken for the object to fall in seconds

The result *h* is the distance or the height in meters.

#### Example 2

Solve \begin{align*}x^{2}-5x+6=0\end{align*} using the "Factor to Solve" method.

The factor method is based on writing the quadratic equation in *factored form*. That is, as a product of two linear expressions. So our equation maybe solved by the following way:

\begin{align*}x^{2}-5x+6 & = 0\\ (x-3)(x-2) & = 0\end{align*}

Recall, that

\begin{align*}a \cdot b=0 \ \text{if and only if} \ a=0 \ \text{and} \ b=0\end{align*}

This tells us that

\begin{align*}x-3=0\end{align*}

or

\begin{align*}x-2=0\end{align*}

which give the roots (or *zeros*)

\begin{align*}x=3\end{align*}

and

\begin{align*}x=2\end{align*}

In other words, the solution set is {2, 3}.

#### Example 3

Solve \begin{align*}x^{2}-5x+6=0\end{align*} by the "Completing the Square" method.

To solve the above equation by completing the square, first move the “\begin{align*}c\end{align*}” term to the other side of the equation,

\begin{align*}x^{2}-5x=-6\end{align*}

Next, make the left-hand side a “perfect square” by adding the appropriate number. To do so, take one-half of the coefficient of \begin{align*}x\end{align*} (the \begin{align*}b\end{align*} coefficient) and square it and then add the result to both sides of the equation:

\begin{align*}b & = -5 \\ \frac{b}{2} & = \frac{-5}{2}\\ \left ( \frac{b}{2} \right ) ^2 & = \frac{25}{4}\end{align*}

Adding to both sides of the equation,

\begin{align*}x^{2}-5x+\frac{25}{4} & = -6 + \frac{25}{4}\\ x^2 -5x + \frac{25}{4} & = \frac{1}{4}\\ \left ( x-\frac{5}{2} \right )^2 & = \frac{1}{4}\end{align*}

this last equation can be easily solved by taking the square root of both sides,

\begin{align*}\left(x-\frac{5}{2}\right)=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}\end{align*}

Hence

\begin{align*}x=\pm\frac{1}{2}+\frac{5}{2}\end{align*}

and the solutions are

\begin{align*}x=3\end{align*}

and

\begin{align*}x=2\end{align*}

which are identical to answers of the factor method.

#### Example 4

Solve \begin{align*}x^{2}-5x+6=0\end{align*} using the Quadratic Formula.

The quadratic formula works for finding the *x*-intercepts in all quadratic equations, therefore it is highly encouraged that you memorize the formula. The other methods can be much faster though, so it is well worth understanding each of the different methods.

Recall, if \begin{align*}ax^{2}+bx+c=0\end{align*} where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} are real numbers and \begin{align*}a\ne0\end{align*}, then the roots of the equation can be determined by the quadratic formula.

\begin{align*}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.\end{align*}

Here the coefficients are \begin{align*}a=1, b=-5,\end{align*} and \begin{align*}c=6\end{align*}.

Substituting into the quadratic formula, we get:

\begin{align*}x & = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}\\ & = \frac{5 \pm \sqrt{25-24}}{2}\\ & = \frac{5 \pm 1}{2}\\ & = 3 \ \text{or} \ 2\end{align*}

which is, again, identical to our two solutions above.

#### Example 5

Solve the following equation by factoring: \begin{align*}4x^{2}+7x=2\end{align*}.

\begin{align*}4x^{2} + 7x - 2 = 0\end{align*}

\begin{align*}0 = (4x - 1)(x + 2) \therefore x = \frac{1}{4}, -2\end{align*}

#### Example 6

Solve the following equation by completing the square: \begin{align*}\frac{2}{3}x^{2}-x+\frac{1}{3}=0\end{align*}.

\begin{align*}x^{2}-\frac{3}{2}x+\frac{1}{2}=0\end{align*}

\begin{align*}x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}-\frac{1}{2}\end{align*}

\begin{align*}(x-\frac{3}{4})(x-\frac{3}{4}) = \frac{1}{16}\end{align*}

\begin{align*}x=\frac{1}{2}, 1\end{align*}

#### Example 7

Solve the following equation by using the quadratic formula: \begin{align*}(z+6)^{2}+2z=0\end{align*}.

\begin{align*}z^{2}+14z+36=0 \therefore x = \frac{-14\pm\sqrt{14^{2}-4\cdot1\cdot36}}{2\cdot1}\end{align*}

\begin{align*}x = -7\pm\sqrt{13}\end{align*}

### Review

- List three ways to write a quadratic function.

Solve with the quadratic formula:

- \begin{align*}m^2 - 5m - 14 = 0\end{align*}
- \begin{align*}b^2 - 4b + 4 = 0\end{align*}
- \begin{align*}4b^2 + 8b + 7 = 0\end{align*}
- \begin{align*}2m^2 - 7m - 13 = -10\end{align*}
- \begin{align*}5r^2 = 80\end{align*}
- \begin{align*}k^2 - 31 - 2k = -6 - 3k^2 - 2k\end{align*}

- If you have an equation with a power of 4, explain how you could solve it using the quadratic formula.

Solve by completing the square:

- \begin{align*}2x^2 - 12x + 26 = 10\end{align*}
- \begin{align*}x^2 - 12x + 29 = -3\end{align*}
- \begin{align*}7x^2 - 14x -64 = -8\end{align*}

Solve by the most efficient method:

- \begin{align*}x^4 + 13x^2 + 36 = 0\end{align*}
- \begin{align*}x^4 + 16x^2 -225 = 0\end{align*}
- \begin{align*}\frac{1}{4}x^2 - \frac{1}{3}x + 1 = 0\end{align*}
- \begin{align*}\frac{2}{7}c^2 - \frac{1}{2}c - \frac{3}{14} = 0\end{align*}

In the quadratic formula \begin{align*} b^2 - 4ac\end{align*} is called the discriminant. The values of the discriminant tell us the nature of the solutions or roots of a quadratic equation, \begin{align*}ax^2 + bx + c = 0\end{align*}

- What value(s) of the discriminant result in two unique real solutions?
- What value(s) of the discriminant result in one unique real solutions?
- What value(s) of the discriminant result in two unique imaginary solutions?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.1.