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# 2.2: Equations of Lines and Quadratic Functions

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## 1. Parallel Lines and Perpendicular Lines

We define parallel and perpendicular lines as follows:

Definition: Parallel and Perpendicular Lines

Let two lines be given with slopes $$m_1$$ and $$m_2$$.

1. The lines are parallel if they have the same slope, that is

$m_1=m_2.$

2. The lines are perpendicular if

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

Example 1

Find the equation of the line that passes through the point $$(1,4)$$ and is perpendicular to the line

$3x - 2y = 5.$

Solution:

First notice that to find the equation of a line we need a point and a slope. We have the point, namely $$(1,4)$$. To get the slope we solve for $$y$$ in the other line:

\begin{align} 2y &= 3x - 5 \\ y &= \dfrac{3}{2}x - \dfrac{5}{2} \end{align}

Since the slope of the perpendicular line is $$\frac{3}{2}$$, the slope of our line is -$$\frac{2}{3}$$. Now we use the point-slope form:

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

Note: Horizontal and Vertical Lines

Recall that a horizontal line is of the form

$y = constant$

and a vertical line is of the form

$x = constant.$

Example 2

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

http://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:

Definition: Standard form of a Quadratic

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

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.

Example 3

Put the following in standard form

$y = 2x^2- 8x + 2$

Solution

1. Factor the leading coefficient:

$y = 2(x^2 - 4x + 1)$

2. Calculate $$-\frac{b}{2}$$

$-\dfrac{b}{2}= -\dfrac{(-4)}{2}= 2$

3. Square the solution above:

$2^2 = 4$

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

$y = 2(x^2- 4x + 4 - 4 + 1)$

5. Regroup:

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

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

$y = 2[(x - 2)^2- 3]$

7. Multiply out the outer constant:

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

## 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

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

Example

If you throw a basketball from five feet with an initial velocity of 25 feet per second, how long will it take to hit the ground? How high will the ball go and when is this maximum height attained? For what times will the ball be higher than 7 feet?

Solution

We have that

$v_0=25 \; \text{ and } s_0=5$

so that

$s = -16t^2 + 25t + 5$

When it hits the ground

$s = 0$

or

$0 = -16t^2 + 25t + 5$

We get the two solutions

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

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.