# What's the Solution the equations below?

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Can you solve this problem?

1. Lines can be used to approximate a wide variety of functions; often a function can be described using many lines.
If a stock price goes from \$10 to \$12 from January 1st to January 31, and from \$12 to \$9 from February 1st to February 28th, is the price change from \$10 to \$9 a straight line?
How can I use two “pieces” of lines to describe the price movements from the beginning of January to the end of February?

2.
1. What is the equation of the line through (1,2) and (3, -1)?
2. What is the vertex of f(x) = x2 – 6x + 3?
3. Using points, straight lines, and possibly the curve of a parabola, construct a piecewise function whose graph looks like a smiling face.

posted Feb 13

### 1.

On of the very best examples of a plot with many, many line segments is the Stock Market prices.

This line has two segments and is not a straight line.

In math, we write multiple equations and then specify when they should be used. For example:
For phone calls:
10 cents/min (midnight to 8 a.m.)
15 cents/min (8 a.m. to midnight)

For this problem:
Stock price is:
\$10 + ((12-10)/31)*day (for January days)
\$12 - ((-9-12)/28)*day (for February days)

### 2.

1) This equation will be in the form y=mx+b , where m is slope and b is y-intercept.

First, we need to find the slope between the two points.

m = (y2 - y1) / (x2 - x1)
m = (-1 - 2) / (3 - 1)
m = -3 / 2

The equation of the line so far is

y = (-3/2)x + b

Now we plug in the values of the coordinate point (1, 2) into the equation to find b.

2 = (-3 / 2)(1) + b
2 = (-3 / 2) + b
7 / 2 = b

The equation you want is

y = (-3/2)x + (7/2)

2) To find the vertex, we put f(x) into vertex form. Any quadratic function in vertex form is

f(x) = a(x - h)2 + k

where:
a is the coefficient of the x2 term
The coordinate of the vertex is (h, k)

f(x) = (x2 - 6x + 9) - 6
f(x) = (x - 3)(x - 3) - 6
f(x) = (x - 3)2 - 6

h = 3
k = -6

Therefore, the vertex is (3, -6).

3) When drawing a smiling face, we know the feature it has is a mouth and two eyes. The mouth represents a parabola that opens upward or the negative half of a circle. The eyes represents a single point, or a small line. Knowing this fact, we can create a piecewise function for each type of curve to represent different facial features of a smiling face.

We can use the origin as our reference point to start off.

For the mouth, lets use the parabola f(x) = (1/5)x2 - 3. One point of the parabola will be (0, -3). Now we want to pick the endpoints of this parabola. To do this, we can find the x-intercepts. Set f(x) equal to zero and solve for x.

(1 / 5)x2 - 3 = 0

(1 / 5)x2 = 3

x2 = 15

x = -√15 and x = √15

So the domain of this parabolic function is -√15 ≤ x ≤ √15.

Now to draw the eyes, we will draw two short horizontal lines that are symmetrical to each other with reference to the y-axis.

f(x) = 2

The domain of this function on the left of the y-axis is -2 ≤ x ≤ -1. The domain on the right side is 1 ≤ x ≤ 2.
.

f(x) = (1/5)x2 - 3 for -√15 ≤ x ≤ √15

``````                      2     for      -2 ≤ x ≤ -1         and          1 ≤ x ≤ 2
``````

*****From wyzant.com

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