2.2: Equations of Lines and Quadratic Functions

1. Parallel Lines and Perpendicular Lines

We define parallel and perpendicular lines as follows:

$$m_1 = -\dfrac{1}{m_2}.$$

$$\begin{align} y - 4 &= \dfrac{3}{2}(x - 1) \\ y &= \dfrac{3}{2} x +\dfrac{5}{2}. \end{align}$$

$$x = constant.$$

Find the equation of the line through the point $(5,4)$ that is vertical

Solution

We know that the form of the equation is:

$$x = constant$$

Since it passes through $(5,4)$ which as x-coordinate 5 that constant must be 5. We conclude that the equation of the line is

$$x = 5.$$

To interactively investigate equations of lines go to

mathcsjava.emporia.edu/~greenlar/ParPerp/ParPerp.html

2. Quadratic Functions in Standard Form

Recall what the function

$$y = x^2$$

looks like. It is a parabola with vertex at the origin. We can use shifting techniques to graph

$$y= 3(x - 2)^2+ 4.$$

We see that the vertex is shifted to the right by 2 units and up 4 units. There is vertical stretching by a factor of three. In general, we say that a quadratic is in standard form if it looks like:

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

Here the $h$ represents the horizontal shift, the $k$ represents the vertical shift, and the $a$ represents the stretching factor. If $a$ is negative the parabola is concave down (looks like a frown).

To put a quadratic in standard form, we complete the square.

$$y = 2(x - 2)^2- 6.$$

3. Applications:

Newton's law of motion:

Newton discovered that if an object is thrown from an initial height of $s_0$ feet with an initial velocity of $v_0$ feet per second then the position of the function of the object is

$$s = -16t^2+ v_0t + s_0$$

$$t = 1.74 \; \text{ and } \; t = -1.74$$

Notice that a negative time means is not the solution, hence the ball hits the ground after 1.74 seconds.

To find out how high the ball will go, we are after the $s$ coordinate of the vertex. We put the equation into standard form

Now we can say that the ball will reach a height of 9.77 feet, $\frac{25}{32}$ seconds after it was released.

To find out when the ball will be higher than 7 feet we solve

$$7 < -16t^2 + 25t + 5$$

or

$$0 < -16t^2 + 25t - 2$$

Using the quadratic formula we find the roots at

$$t = 0.085 \;\;\; \text{and} \;\;\; t = 1.48$$

So that the ball will be higher than 7 feet between 0.085 seconds and 1.48 seconds.

1. Factor the leading coefficient:

$$y = -16(x^2 - \dfrac{25}{16} x - \dfrac{5}{16})$$

2. Calculate

$$\begin{align} -\dfrac{b}{2}&=-(\dfrac{25}{16}) \\ &=-\dfrac{25}{32} \end{align}$$

3. Square the solution above:

$$(-\dfrac{25}{32})^2 = 0.61$$

4. Add and subtract answer from part three inside parentheses:

$$y = -16(x^2 - \dfrac{25}{16} x + 0.61 - 0.61 + 5/16)$$

5. Regroup:

$$y = -16[(x^2 - \dfrac{25}{16}x + 0.61 ) - 0.61]$$

6. Factor the inner parentheses using part two as a hint:

$$y = -16[(x - \dfrac{25}{32})^2 - 0.61]$$

7. Multiply out the outer constant:

$$y = -16(x - \dfrac{25}{32})^2 + 9.77$$

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.