Lecture: Quadratic Functions & Their Graphs
Lecture script · about 15–20 minutes spoken · Week 9
Opening
Think about the top of a flight. A sports analyst wants the highest point of a kicked ball. A small-business owner wants the price that brings in the most money. A gardener wants the biggest plot a fixed length of fencing can wrap around. Each of them is asking the same question: how high, how much, how big. In this lecture you'll learn to answer it from a formula, and then check it on a graph.
Here's where we are in this course. In Week 3 you met functions and their range, the set of all outputs. In Week 4 you found intercepts by setting one variable to zero. In Week 7 you solved quadratic equations, and you found that x squared minus 6x plus 5 equals zero when x is 1 or 5. In this lecture we stop asking only where that expression equals zero and look at every output at once. Next week, in Week 10, Polynomial & Rational Functions, you'll read graphs of higher-degree polynomials the same way.
I'll take eight ideas in order: the shape, its turning point, vertex form, up or down, the vertex formula, the intercepts, switching forms, and maximum and minimum problems. Most examples come from one parabola, with a launched object and a fenced garden along the way.
1. Meet the parabola: a table that turns around
Picture a skateboarder dropping into a half-pipe. The board rolls down, slows through the lowest point, and climbs the far side, and the way up mirrors the way down. Hold that picture, because a table of numbers can do the same thing.
Take the week's parabola, f of x equals x squared minus 6x plus 5. Put in 2. Two squared is 4. Six times two is 12. Then 4 minus 12 plus 5 is negative 3. Do that for every whole number from 0 to 6, and the outputs come out 5, 0, negative 3, negative 4, negative 3, 0, 5. They fall, bottom out at negative 4, and climb back to 5.
A straight line can't do that. A line changes by the same amount at every step, so it can only go one way. This new kind of rule has a name. A quadratic function is a function that can be written f of x equals a x squared plus b x plus c, with a not equal to zero. Its highest power of the input is 2, and its graph is a parabola. And a parabola is the U-shaped curve that is the graph of a quadratic function; it opens either upward or downward and is the same on both sides of a vertical mirror line.
The shape is smooth, too, with no corner at the bottom, only a gentle turn.
Pause and try this. The seven outputs of the week's parabola are 5, 0, negative 3, negative 4, negative 3, 0, 5. What do they add up to?
Add them up before you read on.
Here's the answer. They add up to 0, because the two 5s cancel the negative 4 and the two negative 3s, and the zeros add nothing.
2. Find the turning point: the vertex and the axis of symmetry
Fold a sheet of paper down the middle, and the two halves match. Every parabola has a fold line like that, and you can find it in the table.
Look at the outputs in pairs. The inputs 2 and 4 both give negative 3. The inputs 1 and 5 both give 0. The inputs 0 and 6 both give 5. Every pair sits the same distance from the input 3, one on the left and one on the right. And at 3 itself, the output is negative 4, the lowest value in the table.
So that point is special. The vertex is the turning point of a parabola, written h comma k: the lowest point when the parabola opens up and the highest point when it opens down. For the week's parabola, the vertex is the point 3 comma negative 4. The fold line has a name too. The axis of symmetry is the vertical line x equals h through the vertex of a parabola; points the same horizontal distance from it on either side have equal outputs. Here the axis is the line x equals 3.
Symmetry saves you work: the point 0 comma 5 has a mirror partner three steps right of the axis, at 6, with the same output.
One slip to watch. The axis is a line, so write it as an equation: x equals 3. Not the number 3 on its own, and not the vertex point.
Pause and try this. On the week's parabola, the input 1 gives an output of 0. Using only the axis x equals 3, which other input gives 0?
Think about the mirror before you read on.
Here's the answer. The input 5 gives 0, because 1 sits two steps left of the axis and 5 sits two steps right of it.
3. Read vertex form: the vertex in plain sight
Some ways of writing a function hide the vertex, and one way puts it on display.
Vertex form is a quadratic function written f of x equals a times the quantity x minus h, squared, plus k. You read the vertex h comma k and the axis x equals h off directly, with h taken from inside the parentheses with its sign reversed.
Here's why it works. A square is never negative, so the squared part is smallest when it equals zero, and that happens at exactly one input: x equals h. At that input the output is k.
Take the quantity x minus 2, squared, plus 3. Inside the parentheses it says x minus 2, so h is 2, and k is 3. The vertex is 2 comma 3, and the axis is x equals 2.
Now take the quantity x plus 3, squared, minus 5. Here's the trap. It's tempting to say the vertex is 3 comma negative 5. But x plus 3 is the same as x minus negative 3, so h is negative 3. The vertex is negative 3 comma negative 5. You can test it. Put negative 3 in, and the inside is zero, so the output is negative 5. Put 3 in, and the inside is 6, so you get 36 minus 5, which is 31. That's nowhere near negative 5.
Pause and try this. In vertex form, the week's parabola is the quantity x minus 3, squared, minus 4. Read off its vertex.
Say the point out loud first.
Here's the answer. The vertex is 3 comma negative 4, because x minus 3 means h is 3 and the constant outside is k, negative 4, which matches the table.
4. Decide up or down: minimum, maximum, and range
A satellite dish cups upward, and an umbrella cups downward. A parabola does one or the other, and the number a, in front of the square, tells you which.
When a is positive, the parabola opens up, and the vertex is the bottom. The minimum value is the smallest output a function reaches; for a parabola that opens up it is k, the y-value of the vertex. When a is negative, the parabola opens down, and the vertex is the top. The maximum value is the largest output a function reaches; for a parabola that opens down it is k.
For the week's parabola, a is 1, a positive number, so it opens up and negative 4 is its minimum. That also gives you the range, an idea from Week 3. Every output is negative 4 or more, so the range runs from negative 4, included, up to infinity.
Now try a parabola that opens down: negative x squared plus 4x plus 1. Its a is negative 1. Its vertex, which you'll find in a moment, is 2 comma 5. Because a is negative, 5 is a maximum, not a minimum. The range is everything from negative infinity up to 5, with 5 included.
The common slip is calling every vertex a minimum, because the first parabolas you meet open up. Check the sign of a first.
Pause and try this. The launched object's height has a equal to negative 16. Is its vertex a minimum or a maximum?
Decide before you read on.
Here's the answer. It's a maximum, because negative 16 is negative, so the parabola opens down and the vertex is its highest point.
5. Find the vertex from standard form
Most formulas don't arrive in vertex form. They arrive as a x squared plus b x plus c. That's standard form of a quadratic function: f of x equals a x squared plus b x plus c, with a not equal to zero. You read the y-intercept off, and you find the vertex from a and b.
The tool is the vertex formula: the rule h equals negative b divided by 2a gives the x-value of the vertex from standard form. The y-value is then k equals f of h.
Let's run it on the week's parabola. Read a, b, and c with their signs: a is 1, b is negative 6, c is 5. Negative b is negative of negative 6, which is 6. Two a is 2. Six divided by two is three, so h is 3. Then k is f of 3: 9 minus 18 plus 5, which is negative 4. The vertex is 3 comma negative 4, the same point the table showed.
Now the one that opens down: negative x squared plus 4x plus 1. Here a is negative 1 and b is 4. Negative b is negative 4. Two a is negative 2. Negative 4 divided by negative 2 is 2. Then f of 2 is negative 4, plus 8, plus 1, which is 5. The vertex is 2 comma 5, a maximum.
Watch the double negative. If you drop it and write h as negative 3 for the week's parabola, then f of negative 3 is 32, far from the lowest value in the table. That's your signal to go back.
Pause and try this. For the week's parabola, which number do you plug into the function to find k?
Name the input before you read on.
Here's the answer. You plug in 3, the value of h, because k is always the output at the vertex's x-value.
6. Find the intercepts and let the discriminant predict them
Think of a road map that marks where a highway crosses each border. Intercepts are those crossing points on a graph.
The y-intercept is where the graph crosses the y-axis, so you set x to zero. In standard form every term with an x disappears, and the output is c. For the week's parabola, f of 0 is 5, so the y-intercept is the point 0 comma 5.
The x-intercepts are where the output is zero. That's exactly the Week 7 equation. And here's the nice part: the discriminant, b squared minus 4 a c, tells you how many there are before you solve. For the week's parabola it's 36 minus 20, which is 16. A positive discriminant means two crossings. Factoring gives x equals 1 or x equals 5, so the x-intercepts are 1 comma 0 and 5 comma 0. And the axis, x equals 3, sits exactly halfway between them.
What about a negative discriminant? Take the quantity x minus 2, squared, plus 3. Expanded, it's x squared minus 4x plus 7, and its discriminant is 16 minus 28, which is negative 12. Negative means no crossings. The picture agrees: the parabola opens up from a lowest point at height 3, already above the x-axis, so it never comes down to zero.
A discriminant of zero means one touch: the vertex sits right on the x-axis.
Pause and try this. A parabola's discriminant is negative 12. How many x-intercepts does it have?
Decide before you read on.
Here's the answer. It has none, because a negative discriminant means no real number makes the output zero.
7. Convert between the forms
A recipe can be written as steps or as a list of ingredients, and each version answers a different question fast. Vertex form shows the vertex. Standard form shows the y-intercept and the a, b, and c the discriminant needs. So it pays to switch.
To go from vertex form to standard form, you expand. Take 2 times the quantity x minus 1, squared, minus 3. Square first: x minus 1, squared, is x squared minus 2x plus 1. Then distribute the 2: 2x squared minus 4x plus 2. Then combine the constants: 2 minus 3 is negative 1. So you get 2x squared minus 4x minus 1. Check at zero: the vertex form gives 2 minus 3, which is negative 1, and the standard form gives negative 1 too.
The slip here is distributing before squaring, or squaring only two of the terms. Square first, then distribute, and remember that the quantity x minus 1, squared, has three terms.
To go the other way, you complete the square, the move from Week 7. For x squared minus 6x plus 5, take half of negative 6, which is negative 3, and square it to get 9. Add 9 and subtract 9, which changes nothing. Now x squared minus 6x plus 9 folds into the quantity x minus 3, squared, and negative 9 plus 5 is negative 4. So the week's parabola is the quantity x minus 3, squared, minus 4.
Pause and try this. After expanding, the function 2x squared minus 4x minus 1 came out of a vertex form with vertex 1 comma negative 3. What does the vertex formula give for h?
Run the formula before you read on.
Here's the answer. It gives 1, because negative of negative 4 is 4, and 4 divided by 2 times 2, which is 4, equals 1.
8. Solve maximum and minimum problems
Now the payoff. A question that says how high, how much, or how big is often a vertex question in disguise. The plan: write the quantity as a quadratic function, find the vertex, and say what each coordinate means.
First, the launched object. Its height in feet, t seconds after launch, is negative 16 t squared plus 64 t plus 80. The 16 is half of gravity's pull of 32 feet per second per second. A note on letters: here h names the height function, a different job from the vertex's h. The a is negative 16, so the curve opens down and the vertex is the top. When does it peak? Negative 64 divided by 2 times negative 16, which is negative 64 divided by negative 32, gives 2 seconds. How high? Negative 16 times 4 is negative 64, plus 128, plus 80, gives 144 feet. So the object reaches its greatest height, 144 feet, two seconds after launch. The starting height, 80 feet, is the y-intercept: the platform. This model ignores air resistance, so it describes an idealized launch.
Second, the garden. You have 40 feet of fencing for a rectangle. Call the width w. Two widths and two lengths use 40 feet, so each length is 20 minus w, and the area is w times 20 minus w. That's negative w squared plus 20w. The a is negative 1, so there's a maximum. The best width is negative 20 divided by negative 2, which is 10 feet. The area is 10 times 10, which is 100 square feet: a square. And 8 feet wide gives 96, the same as 12 feet wide, both less than 100.
Pause and try this. For the launched object, someone answers "2" to the question "how high does it go?" What's missing?
Find the gap before you read on.
Here's the answer. The height is missing, because 2 is the time in seconds, and the height at that time is h of 2, which is 144 feet.
Closing
Let's pull it together, in the order we took it. A quadratic function's graph is a parabola, one smooth curve whose table falls and rises. The vertex is its turning point, and the axis of symmetry is the vertical line through it. Vertex form hands you the vertex, with the sign inside the parentheses flipped. A positive a opens up to a minimum, and a negative a opens down to a maximum, and the vertex sets the range. From standard form, h is negative b over 2a, and k is the output at h. The y-intercept is c, and the discriminant counts the x-intercepts before you find them. You expand to reach standard form and complete the square to reach vertex form. And a how-high or how-big question is answered by both coordinates of the vertex, said in words with units.
Next in this module, you'll find the slide deck, then the Readings & Resources page, the lecture tutorial, the practice set, and the quiz. In Week 10, Polynomial & Rational Functions, you'll read the shapes of higher-degree polynomials and fractions of polynomials from their formulas.
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