Learning Objectives
Problem 1 of 3
f(x) = 2x2 − 5x − 12.
(a) Use the method of “completing the square” to place the given quadratic equation in standard form.
(b) Find the root(s) of the quadratic.
(c) State the domain and range in interval notation.
(d) Sketch the graph. Include and label:
- all intercepts
- the vertex
- the axis of symmetry; label it with it's equation
Solution
(a)
i) First we must factor the coefficient of x2 out of the first two terms. Remember to leave some "space" in the brackets
f(x) = 2(x2 − 5/2x) − 12
ii) Divide the coefficent of the x-term by 2 and square the result.
(5/2)/2 = (5/2) * (1/2)
= 5/4
(5/4)2 = 25/16
iii) Now add and subtract the result from step ii) inside the brackets hey we just completed the square!
f(x) = 2(x2 − 5/2x + (25/16) - (25/16)) − 12
iv) Factor the first three terms in the brackets into a perfect square
f(x) = 2[(x - 5/4)2 - 25\16)] -12
v) Distribute the number outside the brackets
f(x) = 2(x - 5/4)2 - 50\16 - 12
vi) Simplify. Change 12 into a fraction that has a denominator of 16
f(x) = 2(x - 5/4)2 - 50\16 - 192/16
vii) Combine Like terms and there you have it the standard form.
f(x) = 2(x - 5/4)2 -242/16
OR to further simplify it:
f(x) = 2(x - 5/4)2 -121/8
(b)
to find the quadratics roots we must use the quadratic root formula:__
(i) sub in the numbers
x= -(-5)± √(5)²-4(2)(-12) / 2(2)
(ii) solve the equation of what is squared
x= 5±√25 + 96 / 4
(iii) simplify what's inside the square root
x= (5±√121 )/ 4
(iv) simplify everything else and there you have it, the roots!
x= (5± 11) / 4
x1= 4
x2= -3/2
(c) Domain: (-∞, ∞)
Reminder! The domain of every parabola is (-∞, ∞)
Range: [-121/8, ∞)
Flashback: f(x) = 2x2 − 5x − 12
f(x) = 2(x-5/4)2 -121/8
Since Parameter a (2) is positive we know the parabola opens up
Therefore.. we know the parabola has a minimum value
Since the K value of the vertex (h, k) is -121/8, we know thats the min. value for y
(d) (Insert Image)
Y-Intercept: (0, -12)
X-Intercept(s) a.k.a. Roots a.k.a. Zeros: (4, 0) (-3/2, 0)
Vertex: (5/4, -121/8)
Axis of Symetry: x=5/4
Problem 2 of 3
When bicycles are sold for $280 each, a cycle store can sell 80 in a season. For every $10 increase in the price, the number sold drops by 3.
(a) Represent the sales revenue as a quadratic function of either the number sold or the price.
Let X be number of increase price
(Sale Price - Increase PriceX)(Units sold in season - units droppedX)
($280 - $10x)(80-3x)
22400 - 800x - 840x + 30x²
30x² - 1640x + 22400 = y
(b) What is the number sold and the price if the total sales revenue is exactly $20 000?
(c) What is the range of prices that will give a sales revenue that exceeds $15 000.
Solution
Problem 3 of 3
coming soon ...
Solution
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