Quadratic relationships
Quadratic relationships are those in which the highest degree (largest exponent or sum of exponents in any of the monomials that make up the polynomial) of the expression is 2.
The parabola
The graph of the quadratic equation f(x) = ax2 + bx + c (a ≠ 0) is a parabola. The parabola is symmetric about its axis of symmetry. The equation of the axis of symmetry is 
and the intersection of the axis of symmetry and the parabola is the vertex, or turning point, of the parabola. If the quadratic coefficient is a > 0, the graph is concave up (opens up), and if a < 0, the graph is concave down.
If the parabola is concave up, the y-coordinate of the vertex represents the minimum value of the range of the function, and if the graph is concave down, the y-coordinate of the vertex represents the maximum value of the range of the function.
The x-intercepts of a graph correspond exactly to the roots of the equation f(x) = 0.
The form f(x) = ax2 + bx + c is called the standard form of the parabola. Another form of the parabola is called the vertex form and is written as f(x) = a(x − h)2 + k. The vertex of this parabola is (h, k).
Converting an equation from vertex form to standard form merely requires that the expression a(x − h)2 + k be expanded and like terms combined. Converting from standard form to vertex form requires a technique called completing the square that will be covered later in this chapter.
Given the parabola f(x) = 3x2 + 12x – 5. Use this function to answer questions 1–3.
1. Find the equation of the axis of symmetry.
2. Find the coordinates of the vertex.
3. Determine the range of the function.
Given the parabola 
, use this function to answer questions 4–6.
4. Find the equation of the axis of symmetry.
5. Find the coordinates of the vertex.
6. Determine the range of the function.
7. Determine the coordinates of the x-intercepts of the parabola q(x) = x2 − 11x + 30.
8. Determine the equation of the parabola in the accompanying image.
Given the parabola p(x) = 2(x + 3)2 − 1, use this function to answer questions 9 and 10.
9. Determine the coordinates of the vertex.
10. Write the equation for the axis of symmetry.
Special factoring formulas
The process for factoring is made easier if you know some basic, frequently used factoring patterns.
1. Common factors: The first step should always be to look at the expression. The common factor may be a constant (as the 2 is in 2x2 + 4x − 6), a monomial (as 3x is in 6x3 + 9x2 − 12x), or even a binomial (as 2x + 3 is in 4x2 (2x + 3) − 11(2x + 3)).
2. Difference of squares: a2 − b2 = (a + b)(a − b).
3. Square trinomials: a2 + 2ab + b2 = (a + b)2 and a2− 2ab + b2 = (a − b)2. Notice that the first and last terms are squares and that the middle term is twice the product of the square roots of the first and last terms. The sign of the middle term agrees with the sign inside the parentheses of the factored form.
4. Difference and sum of cubes: a3 − b3 = (a − b)(a2 + ab + b2) and a3 + b3 = (a + b)(a2 − ab + b2). You need to be careful with these formulas because they look very similar to the square trinomials. Make note of the differences: the sign between the terms in the binomial is the same as the sign between the original cubes, and the sign of the middle term in the factor is the opposite of this; the coefficient of the middle term in the trinomial factor is 1 (not 2, as it is in the square trinomial terms).
Completely factor each of the following.
1. 8x4 − 12x3 + 20x2
2. 8x2(4x − 3) + 3x(4x − 3) − 9(4x − 3)
3. 100x2 + 140xy + 49y2
4. 100w2 − 100wv + 25v2
5. 81t2 − 144
6. (3x + 4)2 − (2y − 1)2
7. 16x3 + 54
8. 216 − 125k3
Trial and error
Perhaps a good way to introduce the notion of the trial and error method of factoring is to look at two multiplication problems.
There are some basic arithmetic facts that can assist in the solution of this kind of problem. There are also some approaches that use technology, one of which will be examined in the next section.
Arithmetic approach
The simple concepts that a negative product comes from the multiplication of numbers with different signs, and that an odd sum comes from adding an even and an odd number, can help eliminate some of the multitude of factors that can arise in a problem like 60x2 + 103x − 72. The third term constant is negative, so you know that the factors of 72 shown in the parentheses in the previous solution will have opposite signs, and that 103 came from the difference rather than the sum of terms. Because the middle term is odd, there must be an odd product and an even product from the multiplication of the inner and outer terms of the original binomials. As a result, 2 and 30, and 6 and 10 can be eliminated as candidates for the factors of 60, and 2 and 36, 4 and 18, and 6 and 12 can be eliminated as factors of 72. The original problem has no common factors so neither of the binomial factors can have a common factor. 3 and 24 can be eliminated as candidates for 72. 72 − 60 does not equal 103 so the factors 1 and 60 and 1 and 72 can be eliminated from consideration. This leaves 3 and 20, 4 and 15, and 5 and 12 as the possible factors of 60; and 8 and 9 as the factors of 72. Check (the trial aspect) the various products of the factors of 60 with 8 and 9. (9)(15) = 135 and (4)(8) = 32 and the difference between them is 103. Therefore, the factors of 60x2 + 103x − 72 are 4x + 9 and 15x − 8.
Graphical approach
Use a graphing utility to sketch the graph of f(x) = 60x2 + 103x − 72.
Since it is only the x-intercepts that need to be seen, the viewing window does not need to be set to see the vertex of the parabola.
Use the zero feature of your calculator to find the x-intercepts of the graph. The image below was created using a TI-Nspire. The intercepts are associated with the variables x1 and x2. Use the approxRational command to convert these numbers to fractions. The command parameters allow for a tolerance level, and in these cases, 0.005 is chosen.
The zeroes of the function are 
. To get the factors of the original quadratic, manipulate each of these equations back to binomials.
Factor each of the expressions.
1. 12x2 + 13x − 55
2. 48x2 + 110x + 63
3. 24x2 + 22x − 65
4. 36x2 + 69x + 28
5. 50x2 − 105x + 54
6. 54x2 + 69x + 20
7. 48x2 + 8y − 225
8. 24x2 + 35x − 150
Completing the square
The “square” in the process of completing the square is the square trinomial (which will have to be square if one chooses to provide a geometric model of the algebraic expression). Recall that a2 + 2ab + b2 = (a + b)2 and a2 − 2ab + b2 = (a − b)2. The middle terms are twice the square roots of the first and third terms. This is the key for completing the square: Take one half the coefficient of the linear term of the expression, square it, and add it.
The process is a little more complicated if the quadratic coefficient is not 1. In that case, first step is to factor the quadratic coefficient from the quadratic and linear terms and then proceed as usual.
Completing the square can also be used to solve quadratic equations that cannot be factored.
Rewrite the equations for each parabola in questions 1−4 in vertex form.
1. f(x) = x2 − 6x + 5
2. g(x) = 5x2 + 15x − 3
3. 
4. 
Use completing the square to solve each of the equations in questions 5−8.
5. x2 + 8x − 5 = 0
6. 4x2 + 8x − 5 = 0
7. 
8. 6x2 + 13x + 6 = 0
Quadratic formula
When applied to the general quadratic equation ax2 + bx + c = 0, the process of completing the square yields the quadratic formula.
Use the quadratic formula to solve these equations.
1. 4x2 − 9x − 3 = 0
2. −16x2 + 128x + 56 = 0
3. 
4. 
5. −4.9x2 + 14.3x + 5.8 = 0
Applications
When an object is thrown vertically into the air from the top of some height, the position of the object can be computed by the formula:
where g represents the acceleration constant due to gravity (32 ft/sec2 in standard units; 9.8 m/sec2 in metric units), v0 is the initial velocity of the object, and s0 is the initial height of the object.
Solve each of the following problems.
A stone is thrown vertically into the air from the top of a cliff 40 ft above the ground with an initial upward velocity of 64 ft/sec.
1. What is the maximum height of the stone?
2. When is the stone at a height of 75 ft above the ground?
3. When does the stone hit the ground?
A ball is thrown vertically into the air from the roof of a building 15-m tall with an initial upward velocity of 19.6 m/sec.
4. What is the maximum height of the ball?
5. When is the ball at a height of 30 m?
6. When does the ball strike the ground?
A business determines that its revenue function for selling x items of its product is given by the equation R(x) = −x2 + 1200x and that the cost of producing these x items is given by the function C(x) = 100x + 15,000. Revenue and cost in dollars.
7. How many items should the company sell to maximize its revenue?
8. What is the maximum revenue?
9. How much does it cost to produce these items?
10. Profit is the difference between the revenue and cost. What is the profit for producing and selling these items?
11. Write an equation for the profit function.
12. At what point does the company begin to make a profit? When is the company working with a deficit (losing money)?
13. How many items must be produced and sold to maximize the profit?
14. What is the maximum profit?
Square root function
The graph of the equation y = x2 passes the vertical line test (y = x2 is a function) but fails the horizontal line test (the inverse of y = x2 is not a function).
Yet, it is known that the square root function is the inverse of the function y = x2. The explanation for this contradiction involves the important technique of restricting the domain. If the domain of the parabola y = x2 is restricted to x > 0, the right-hand branch of the parabola passes both the vertical and horizontal line tests (making it a one-to-one function) and its inverse is the square root function.
Determine the restricted domain of the given equations so that the function is one to one and provide the equation of the inverse function.
1. 
2. y = −3(x − 5)2 + 2
Circles
By definition, a circle is the set of all points in a plane at a fixed distance r from a fixed point (h, k). Translating this definition into an equation, a point (x, y) is on the circle if its distance from (h, k) is equal to r, or 
Square both sides of the equation to get (x − h)2 + (y − k)2 = r2. Of course, r is the length of the radius of the circle, and (h, k) are the coordinates for the center of the circle.
Solve the following problem.
1. Find the equation of the circle centered at (−5, 2), which passes through the point (4, 3).
Find the coordinates of the center and length of the radius for the circles whose equations are given in questions 2−5.
2. (x + 4)2 + (y − 9)2 = 81
3. (x − 6)2 + y2 = 18
4. x2 + y2 + 14x − 20y − 20 = 0
5. x2 + y2− 42x + 120y − 1000 = 0
Ellipses
The algebraic definition of an ellipse is the set of all points in a plane with the property that the sum of the distances from two fixed points is a constant. The geometric definition is that an ellipse is the intersection of a plane and a cone in which the plane is at an angle between being parallel to an edge of the cone and parallel to the base of a cone. A good example of an ellipse is to hold a cup of water at an angle so that the water does not spill. The surface of the water in the cup will be in the shape of an ellipse.
The following figures show the graphs of an ellipse when the fixed points lie on the x-axis, when they lie on the y-axis, and the equations for each. In each case, the center of the ellipse is the origin.
Observe that the longer axis (called the major axis) of the ellipse in each case has length 2a, while the shorter (minor) axis has length 2b. It is possible to determine the orientation of the major axis by examining the equation. The larger denominator represents the value of a2, and the variable in the numerator reveals the orientation of the major axis.
If the center of the ellipse is translated from the origin to the point (h, k), then the equations of the graphs in standard form become
The ellipse is symmetric with respect to both the major and minor axes.
Solve the problems in this exercise as indicated.
1. The endpoints of the major axis of an ellipse have coordinates (−2, 5) and (−2, −7). If an endpoint of the minor axis has coordinates (−5, −1), write an equation for the ellipse.
2. Find the coordinates of the x- and y-intercepts for an ellipse with the equation 
3. The endpoints of the major axis of an ellipse are located at (2, −3) and (12, −3). If the ellipse is tangent to the x-axis, write an equation for the ellipse.
4. Write, in standard form, the equation for the ellipse defined by 3x2 + 5y2 + 18x − 40y − 43 = 0.
Hyperbolas
The algebraic definition of a hyperbola is the set of all points in a plane with the property that the absolute value of the differences of the distances from two fixed points is a constant. The geometric definition of a hyperbola is the intersection of a plane and both naps of a double-napped cone in which the plane is at an angle between being parallel to an edge of the cone and perpendicular to the bases of the double-napped cones. The outline of the cooling towers in nuclear reactors are in the shape of hyperbolas (the shape of the cooling tower is called a hyperboloid.)
Sketches of two hyperbolas, each centered at the origin, one with the vertices on the x-axis and the other with the vertices on the y-axis, along with their equations in standard form, are shown in the following images.
Observe that the variable in the numerator of the leading fraction indicates the direction in which the vertices of the hyperbola are located from the center. The first denominator is always a2. The rectangle formed has dimensions 2a × 2b, and the diagonals of the rectangle are called the asymptotes. As the absolute value of x gets very large, the graph of the hyperbola becomes much closer to these lines. The asymptotes are very useful for graphing the hyperbolas, and also give an indication that if each of the equations for the hyperbola is solved to be in the form y = , the resulting equations will contain terms in x that become negligible as |x| becomes very large.
If the center of the hyperbola is translated from the origin to (h, k), then the equations for the resulting hyperbolas will be:
There is another equation for a special set of hyperbolas. Equations of the form 
where c is a constant, are hyperbolas that lie in Quadrants I and III when c > 0, and in Quadrants II and IV when c < 0.
Solve the following problems.
1. Determine the coordinates of the center, the coordinates of the vertices, and the equations of the asymptotes of the hyperbola with equation 
2. Determine the coordinates of the center, the coordinates of the vertices, and the equations of the asymptotes of the hyperbola with equation 
3. Rewrite the equation 49y2 − 16x2 + 32x − 686y + 1601 = 0 in the standard form of the hyperbola.
4. Rewrite the equation 81x2 − 64y2 + 324x + 1152y − 10,044 = 0 in the standard form of the hyperbola.
Systems of equations
For the most part, the systems of equations you have been asked to solve up to this point have all involved linear equations. In this section, you will examine linear–quadratic and quadratic–quadratic systems of equations.
