Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.
Chapter 3 · pp. 146–219
This chapter studies quadratic functions, whose graphs are parabolas. You will learn to recognize them in tables, graphs and equations, and to move between standard, factored and vertex forms. You will use these forms to find zeros, vertices, maximum and minimum values, and to solve problems about profit, projectiles, area, inverses, radicals and intersections with lines. This part covers the basic properties of quadratics and how to find their maximum or minimum values.
A quadratic function has degree 2. It can be written in standard form as , where . The term makes the graph curve, and with no domain restrictions the graph is a full parabola. For example, is quadratic because its highest power is 2 and (pp. 150, 155).
A quadratic can be shown as an equation, a table of values or a graph (p. 155).
Pick equally spaced x-values. Subtract consecutive y-values to get the first differences, then subtract those to get the second differences.
A snowboard business has profit values (in -32, -14, 0, 10, 16, 18, 16, 10, 0, -14x=0918, 14, 10, 6, \dots-4$, so it is quadratic (pp. 151, 155).
The sign of the second differences tells you the opening direction. Positive second differences mean the rate of change keeps growing, so the curve bends up and . Negative second differences mean the rate keeps shrinking, so the curve bends down and . The profit data had second differences of , so its parabola opens down (pp. 151, 155).
Vertex form is . The vertex is and the axis of symmetry is . The value of sets the direction (up if positive, down if negative) and how steep the parabola is. For , the vertex is and the parabola opens up because (pp. 151, 154, 155).
Writing vertex form from a graph. Read the vertex off the graph and substitute it for and . Then substitute one other point and solve for . The profit graph has vertex , so . The point gives , so . The model is (p. 151).
Factored form is . The numbers and are the zeros, which are the x-intercepts of the graph. If either factor is 0, the whole product is 0. So has x-intercepts and (pp. 152, 155).
Writing factored form from a graph. Read the zeros, write , then use one other point to find . For the profit graph the zeros are 2 and 8. The point gives , so and . Expanding this gives the same standard form as the vertex-form model, since both describe the same parabola (p. 152).
The axis of symmetry is the vertical line through the vertex. The two halves of the parabola mirror each other across it. When zeros exist, the axis lies midway between them, at (pp. 153, 155).
To find zeros, set and solve. In a real situation a zero can mean that an object lands or a business breaks even.
Example: a tool is tossed with height metres after seconds. Factoring gives . The window is where , so it is 25 m up. Setting gives or . Time cannot be negative, so the tool lands after 5 s. The axis is , and m is the greatest height (p. 153).
With no restrictions, the domain of a quadratic is all real numbers. The range is bounded by the vertex's y-coordinate: if the parabola opens up, if it opens down. Context can restrict both. For the tool toss, the domain is and the range is (pp. 153, 154).
Follow these steps for .
The domain is all real numbers and the range is (p. 154).
The sign of is the same in standard, factored and vertex form. If , the parabola opens up and the function has a minimum. If , it opens down and the function has a maximum. Either way, the extreme value is the y-coordinate of the vertex (pp. 158, 160, 162).
A golf ball's height is metres. Since , it has a maximum. Power lines are 185 m high, so we need the vertex to see whether the ball reaches them.
Method 1: completing the square (always works).
Here: . The vertex is , so the ball peaks at 180 m after 4 s and misses the 185 m lines (pp. 158, 159).
Method 2: factoring (only when zeros exist). Factor to find the zeros, average them for the x-coordinate of the vertex, then substitute to get the y-coordinate. Here has zeros 10 and , which average to 4. Then . Some quadratics have no zeros, so completing the square is the method that always works (pp. 159, 160).
Method 3: a common factor of the terms. This is useful when the full trinomial does not factor easily. Factor only the first two terms to find two points at the same height. Their x-values average to the axis of symmetry. For , we have , which is 0 at and . So and lie on the graph. The axis is and (pp. 161, 162).
Method 4: a graphing calculator. This helps when the coefficients are decimals. Choose a window that shows the vertex, graph the function, and use the minimum or maximum operation. For the steamboat cost , a window with from 0 to 10 works. The calculator gives a minimum cost of $127.70 per hour at about 4 km/h (pp. 160, 161).
The demand function gives the selling price when items are sold. Revenue is demand times the number sold:
Profit is revenue minus cost:
where is the cost of making items. With and , we get and (p. 161).
Maximizing profit.
For the magazine example the vertex is . Here is in thousands of magazines and is in thousands of dollars. So the maximum profit is $6000, reached when 3000 magazines are sold (pp. 161, 162).
An inverse function undoes the original. Every input and output trade places, so the point on the original becomes on the inverse. The inverse answers the question "which input gave this output?" (pp. 165–166)
Example: a cube has surface area . The inverse takes a surface area and returns the side length. If Suzanne has 864 cm² of cardboard, then , so the side is 12 cm. (p. 165)
Swapping coordinates is the same as reflecting the graph in the line . The domain and range also swap. The domain of the inverse is the range of the original, and the range of the inverse is the domain of the original. For , the domain is all real numbers and the range is . So the inverse relation has domain and range all real numbers. The point (2, 4) on becomes (4, 2) on the inverse. (p. 166)
Misconception: the inverse of every quadratic is a function. This is false. If the domain is all real numbers, the parabola fails the horizontal line test. Its inverse then fails the vertical line test, because every except the vertex value has two -values. For the inverse has both (4, 2) and (4, −2). The inverse is a parabola opening right if and left if . (pp. 166, 170)
To find the inverse of a quadratic:
For , swapping gives . Then , so . The root is the upper branch and the root is the lower branch. (pp. 167–168, 170)
On a graphing calculator, enter the two branches separately, one with and one with . (p. 168)
To make the inverse a function, restrict the original domain to one side of the vertex. For with , only the right half of the parabola is used. Then , with a single branch. (pp. 167–168, 170)
Inverses in real situations. When letters have fixed meanings, such as for time and for surface area, do not swap them. Just solve for the input variable. For with , this gives . The original has domain and range . In the new model these swap: the domain is and the range is . (p. 169)
A radical is a root, such as . An entire radical has coefficient 1, like . A mixed radical has another coefficient, like . These two examples are equal values written differently. Like radicals have the same number under the radical sign. For example, and are like radicals, but and are not. (pp. 173, 175)
Multiplying. For and , . More generally, . Multiply the coefficients together and the radicands together. For example, , and . (pp. 173, 175, 177)
Entire to mixed radical. Factor out the greatest perfect square, split the radical, and take the square root of the square. For the greatest perfect square factor is 36, so . Using 4 or 9 would leave a radical that still simplifies. Likewise . (pp. 174, 177)
Mixed to entire radical. Write the coefficient as the square root of its square, then combine under one radical. So . A negative sign stays outside, because a square root of a real number cannot be negative. So . (p. 174)
Exact or approximate. An answer in radical form is exact. A decimal such as is only an approximation. A mixed radical is in simplest form when the number under the radical is as small as possible. That is why is better than . (pp. 173, 177)
Only like radicals combine into one term. Add or subtract the coefficients: . Unlike radicals do not combine. A check: , but . So . (pp. 175, 177)
Simplify every radical to a mixed radical first. This can reveal like radicals that were hidden. Take the perimeter . Here , and . Grouping gives . (p. 176)
To multiply binomials with radicals, expand with the distributive property (FOIL). Then simplify each term and combine like radicals. For example:
Since and , this becomes . (p. 176)
A break-even point is a sales quantity where profit is zero, so revenue equals cost. You find it by solving . On the graph, these are the x-intercepts of the profit function. (pp. 172–173)
Example: , with in thousands of parts and in thousands of dollars. A graphing calculator's zero operation gives and . Anthony breaks even at 4000 parts or at 12 000 parts. (p. 172)
Solving by factoring. Set the expression equal to zero, factor out any common factor, factor the trinomial, and set each factor equal to zero. Here becomes , then . So or . This works only when the expression factors neatly. (p. 173)
The quadratic formula. For any equation ,
For Anthony's equation, , and . Then , which gives or . A profit function does not always have two break-even points. The formula shows whether real solutions exist. (pp. 173–174)
For , draw vertical rays (parallel to the axis of symmetry) that hit the parabola at points such as (4, 4), (2, 1), (−2, 1) and (−4, 4). Each ray reflects so that it makes the same angle with the tangent line as the incoming ray. All the reflected rays cross the y-axis at the same point, (0, 1). This point is the focus. It is why a parabolic satellite dish puts its feed horn at the focus, so the incoming signals gather there. (p. 181)
Three names describe the same numbers. Take a quadratic function and set . This gives the equation . Its solutions are called roots. They are also the zeros of the function and the x-intercepts of its graph (p. 188).
Example: for the roots are and . These are exactly the two points where the parabola crosses the x-axis (p. 188).
Break-even points use the same idea. Profit is zero at a break-even point, so the break-even quantities are the zeros of . For , the zeros are 4 and 5. If is in thousands, the company breaks even at 4000 and at 5000 items (pp. 188, 193).
First write the equation in standard form, . Then pick one of three methods (pp. 188, 210):
When you substitute, keep the signs of , and . For , we have , , . Then , so or (pp. 188, 190–191).
You can leave answers exact, as radicals or fractions. You can also round them to decimals if the problem asks for an approximation (p. 188).
Suppose a balloon's height in metres after seconds is .
Domain and range. Use the zeros to limit the inputs and the vertex to limit the outputs.
Reaching a given height. To find when the balloon is 30 m high, set the function equal to 30. Move everything to one side:
The quadratic formula gives s or s. Both answers are valid. The balloon passes 30 m once on the way up and once on the way down (p. 185).
A quadratic equation often gives two answers, but the situation may allow only one. An inadmissible solution solves the equation but does not make sense in the problem (p. 185). Examples are a negative time or a negative length. Always check each root against the context.
Worked problem: the factory and the lawn. A factory sits on an 80 m by 60 m lot. A lawn strip of uniform width surrounds it, and the lawn's area equals the factory's area. Find the strip's width.
The steps for any area problem are: define a variable, write expressions for the dimensions and areas, set up the equation, solve it, then reject inadmissible roots (pp. 186–187).
A parabola can miss the x-axis, touch it once, or cross it twice. It cannot cross more than twice. So a quadratic function has 0, 1 or 2 zeros (pp. 189, 194).
Using the vertex and the direction of opening. Compare the vertex with the x-axis (pp. 189–190, 194):
To find the vertex, rewrite the function in vertex form by factoring or completing the square (pp. 189–190):
Using factored form. Each factor set to zero gives a zero. For , the zeros are and . If the trinomial is a perfect square, there is only one zero. If it will not factor, use another method (pp. 189–191).
The discriminant is the expression under the radical in the quadratic formula:
You can find it without finishing the formula. For , it is (pp. 190, 192).
Its sign tells you the number of zeros (pp. 190–192, 194):
| Discriminant | Zeros |
|---|---|
| 2 | |
| 1 | |
| 0 |
Here is why. A positive discriminant gives two different values from the . A zero discriminant makes the add nothing, so only one value is left. A negative discriminant needs the square root of a negative number, which is not a real number. So the graph never meets the x-axis.
Examples:
Finding a parameter. To make have exactly one zero, set the discriminant to 0:
Both values give a parabola that touches the x-axis once (p. 193).
Can a business break even? Break-even means . To answer yes or no, you only need the discriminant. For :
There are no zeros, so the business cannot break even in its first year (p. 193).
Some students think every quadratic equation has two solutions. That is not true. It may have two real roots, one repeated root, or no real roots. For example, has discriminant , so it has no real solutions. Check the discriminant to see which case applies (pp. 189, 194, 210).
A family of parabolas is a group of parabolas that share one feature. The shared feature can be the x-intercepts, the vertex, or the y-intercept. Equations in the same family can look very different when they are written in standard form (p. 198).
The feature that stays fixed depends on which form you vary, and the parameter that changes is .
The family idea gives a method. Write the general equation for the known feature, then use one extra point to find (p. 201).
Zeros known. Write , substitute the extra point, and solve for . Take zeros 2 and and the point . Then , so and (pp. 198, 201).
Vertex known. Write and substitute another point. For a vertex and the point : The model is (pp. 200–201).
Example: the overpass. Put one edge of a 10 m wide highway at the origin. The zeros are then 0 and 10, so . The overpass is 13 m high at , so and . The negative means the arch opens downward (p. 199).
To model data with a quadratic, follow these steps (pp. 200–201):
In the smoking data, the percent was 22.6 in both 1989 and 1991. By symmetry the vertex is near the year 1990, so is a sensible estimate. The result only approximates the trend, because the curve was fitted by eye.
A line and a parabola meet in at most two points. The line may cut the parabola twice, touch it once (a tangent line), or miss it completely (pp. 196, 198, 210–211).
Algebraic method. Set the two expressions equal, rewrite as , solve for , then substitute back to find (pp. 195–196).
Example: a skydiver's height is and the parachute-open height is . Equating them gives . The quadratic formula gives s or s. The negative time is inadmissible, so we reject it. Then m (pp. 195–196).
Graphical method. Graph both functions and adjust the window until the crossing shows. Then use the calculator's intersection tool. It is fast, but the answers are approximate. For the skydiver, it gives about (pp. 194–195).
Predicting the number of intersections. Equate the functions and find the discriminant of the resulting quadratic. A positive value means two intersections, zero means one, and negative means none. For and , equating gives . The discriminant is , so the graphs never meet (p. 196).
Skeet shooting. Setting gives . The roots are and , and only the positive one is admissible. The bullet hits at 0.5 s, at height m (p. 197).
Misconception. A line does not always meet a parabola in exactly two places. It can meet it twice, once or never. Even when the algebra gives two solutions, one may not make sense in context, such as a negative time (pp. 196, 198, 210–211).
Each form of a quadratic shows a different feature most easily (pp. 214, 216, 218):
Choose the form that gives the feature you need. For example, break-even points are not clearly visible in vertex form.
Reflecting a parabola in gives a sideways parabola, which fails the vertical line test. So the inverse of a quadratic is not a function unless the domain is restricted to one side of the vertex. For , restricting to makes the inverse a function. That inverse involves a square root (pp. 212, 217–218).
