PAM - Solve Quadratic and Exponential Equations Lesson

Solve Quadratic Equations

In order to solve quadratic equations concepts learned in earlier lessons must be applied.

A parabolic equation always has a vertex, at the midpoint of the equation. The vertex is either the maximum or the minimum of the equation.

The x-intercepts provide solutions to the equation. As seen in the example below, the x-intercepts are sometimes beginning points and ending points.

A formula to use to find x-intercepts when factoring is not an option is the Quadratic Formula. This formula always provides the intercepts and thus factors of the equation.

Quadratic Formula

The key to solving with quadratic equations is understanding the parts of the equation as it relates to the problem.

Example 4: The path of a kicked football is modeled in feet by $y=-0.008x\left(x-147\right)$

How far is the football kicked in feet.
What is the maximum height of the ball?

When the ball hits the ground again, the height is zero. So we are looking for the roots, or x-intercepts.

Rewrite the equation in Intercept Form

y = -0.008(x - 0)(x - 147)

Thus the ball is on the ground at 0 feet, when the kick is made and lands on the ground again at 147 feet. The ball is kicked 147 feet.

Since the equation is parabolic and symmetric and the leading coefficient is negative, the vertex would be the maximum. The vertex would be found at the midway point.

(0 + 147)/2 = 73.5 feet

Plug 73.5 into the equation and solve for y to give the maximum height.

f(73.5) = -0.008(73.5 - 0)(73.5 - 147) = 43.22 feet

The maximum height of the football is 43.22 feet

Creation of Exponential Equations

Now let's examine the exponential equation. Remember this equation is the one that has the asymptote which is not crossed.

Exponential equations are used to provide answers to items having growth or depreciation. An example of growth is your savings account as interest is added for the bank's use of the money for the time that you have it in the account. Depreciation happens after you buy a new car and drive it off the car lot, it is never a new car again.

Bacteria is another real world example of growth or decay. Anything that grows or decays at a fixed rate can be modeled using an exponential function.

y = ab^x + c
where
a = the stretch factor
b = the growth or decay factor, a rate
c = the shift of the horizontal asymptote up or down

c = 0 for these examples

Example 5: Find the exponential equation of the graph if the graph passes through the points (2, 18) and (4, 162).

Step 1. We need to use the equation and create a system of equations to work with by plugging in x and y.

18 = ab² 162 = ab⁴

Method 1: We could solve both for a, set them equal to each other by substitution and solve for b.

$\frac{18}{b^2}=\frac{162}{b^4}$ set the two equation = since both = a

by cross multiplication

division to combine like terms

b ² = 9

b = 3 answer is ±3 but b cannot be negative in an exponential equation, so the final answer is b = 3

Method 2: Create a proportion of like parts.

$\frac{ab^4}{ab^2}=\frac{162}{18}$ start with a proportion of like parts

b² = 9

b = 3 answer is ±3 but b cannot be negative in an exponential equation

Step 2. Solve for a.

18 = ab² 162 = ab⁴

Method 1: Using either of the original equations, solve for "a" first and then plug in b.

a = a =

Method 2: Using either of the original equations, plug in the value of b and solve for "a".

18 = a(3)² 162 = equation image indicator

a = a =

a = 2 a = 2

Step 3: Plug in the "a" value and "b" value to check your work to see if x and y now work (x, y) as an ordered pair. The letter y should be correspond to the values for the ordered pairs (2, 18) and (4, 162).

y = 2(3)^x Check your work by plugging in the x values 2 and 4 to obtain the associated y values.

Watch the Exponential Form video. Take information from a graph or table to create the exponential equation.

Exponential Form Practice

Now that you have watched the video for another example, try a few problems on your own. Check the answer to see if you have the problem correct, check your work, and then if you are still stuck, check the solution provided in this handout Links to an external site..

Solving Exponential Equations

Using the exponential function, we can model growth and decay by allowing the b to be the rate. A rate starts at 100% and either increase (growth) or decreases (decays).

Exponential Growth
y = a (1 + r)^t
where
a is the beginning amount
r is the rate of increase
t = time interval
note: b = (1 + r) for growth

Exponential Decay

y = a (1 - r)^2
where
a is the beginning amount
r is the rate of decrease
t = time interval

Note: b = (1 - r) for decay

Example 6: A study of computer security in small-town A examined the number of incidents reported during a 5 year period. The first year 253 incidents were reported. The growth each year averaged 30%. Write an exponential equation to model the growth in terms of time t in years.

In Example 1 notice that time t is in years since the rate is in years. The rate r and the time t must have the same dimension in terms of time. If they do not, one or the other must have the dimensions adjusted to match.

Now that you have an equation, you can estimate the number of incidents in the town in 8 years with your calculator.

or approximately 2064 incidents

Note: a part of an incident would not be a valid entity.

Example 7: You purchase a car in 2015 for $23,000. The car dealer tells you that the car will depreciate at a rate of 12% yearly.

a. Write an equation to model this information.

r is the percent written as a decimal

t will be the number years

b. If you want to find out how much your car had depreciated in half a year, the formula input would need to be adjusted.

time would be indicated in the number of 1/2 years

r/2 = .12/2 = .06

t is the number of half years (2 per year, so t = 1)

c. What is the value of the car after 1/2 year has passed?

There are 2 periods per year, so t = 1

= $21,620

d. What is the value of the car after 3 years have passed?

Calculation at half years

Calculation at the end of each year

Note that the answers are different for extended periods of time in part d. This is because the value of the car is recalculated more often for the smaller divisions of time. This is called compounding. With smaller recalculation periods, the $23,000 is reduced and recalculated more often at different values.

When you are investing interest rates for savings, the more often the interest is added to the balance and then recalculated again, the more you will gain.

Try a few problems on your own. Check the answer to see if you have the problem correct, check your work, and then if you are still stuck, check the solution provided in the handout Links to an external site..

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