Quadratic functions are usually the first we encounter that have curved or nonlinear graphs. The graph of a quadratic function is a specific kind of curve called a parabola, a sort of Ushaped figure.
All quadratic functions include a term that contains the square of the independent variable, like $x^2.$ There can be no higher power of $x$ in a quadratic function. Some examples are
Quadratic functions have the general form
where A, B and C are fixed numbers (constants) that don't change, and the highest power of the independent variable (x) is 2.
Notice that in $Ax^2 + Bx + C,$ the x in the second term has an exponent of 1; we just aren't in the habit of writing those "invisible" 1's. The parent function of all quadratic functions is the simplest, with A = 1, B = C = 0, f(x) = x^{2}
Not all Ushaped graphs are parabolas. The parabola is the unique graph of a quadratic function.
Quadratic functions are very important. There's a lot to cover, so the material is broken up into sections.
The basic or general form of a quadratic function is shown below, where A, B and C are fixed, numerical constants, and where B or C can be zero. If A = 0, of course, there is no x^{2} term and it's not a quadratic. Terms with x to the first and zero powers are shown, but in practice we write x^{1} = x and x^{0} = 1 (which is not written at all  the ghost 1).
The form is usually written like this,
$$f(x) = Ax^2 + Bx + C$$
and we can find other clever ways to disguise it.
Here are graphs of two parabolas. Use them to learn the terminology we'll use when referring to graphs of curved and parabolic functions.
A parabola is characterized by its vertex (the lowest or highest point of the graph–the bottom or the top), by the direction in which it "opens" – up or down, and by the steepness of the rise of its ends. The ends can rise or fall very steeply or the graph can be very shallow — or anything in between.
The vertex of any parabola is either a global maximum (highest point) or a global minimum (lowest point). It's the point where the slope of the curve (which we define as the slope of the line tangent to the curve at the point of interest) switches from + to  or  to +. The vertex of a parabola can be anywhere in the xy plane.
All parabolas have an axis of symmetry that passes vertically through the vertex. Either side of the parabola is the mirror image of the other across that line. That kind of symmetry is called reflection symmetry.
The roots of a quadratic function (or any function) are the points where it crosses the xaxis, where $f(x) = 0.$ Strictlyspeaking, a quadratic function always has two roots, but they might not be real numbers. A parabola can touch the xaxis just once, as on the right above (a "double root"), twice as on the left (two real roots), or not at all, in which case the roots are "imaginary" and can't be plotted on an xy graph. More on those later.
The function $f(x) = x^2$ is called an even function because $f(x) = f(x);$ for example, $f(2) = f(2) = 4.$
The domain of a quadratic function is always $(\infty, \, \infty;).$
The range of a quadratic function is from the vertex to ±∞, depending on the direction of opening: $(\infty, \, y] \; \text{or} \; [y, \, \infty).$
By applying the horizontal and vertical translation transformations we learned in the functions section, we can derive what we call the vertex form of the quadratic function:
where A is the vertical scaling parameter and (h, k) is the vertex (It's called the vertex form because the coordinates of the vertex embedded in it are obvious). If we expand this (expand means to carry out the squaring of the binomial and distribute the A), we get
Notice the correspondence between the coefficients of the standard form of the quadratic equation: A = A, B = 2Ah and C = Ah^{2} + k. It is from the correspondence for B that we arrive at an easy formula for the xcoordinate of the vertex of a parabola:
xcoordinate of the vertex:
The ycoordinate is then easily found by plugging h in to the function to get y = f(h). Remember that that's what functions are for – you put in an x to get the y that goes with it in (x, y). You should memorize the formula for the xcoordinate of the vertex. You'll need to use it quite a lot to solve problems: h = B/2A.
Video 2 below shows you how to convert between standard and vertext form.
The vertex of a parabola (h, k) is obvious if the function is written in vertex form
and if it's written in standard form, $f(x) = Ax^2 + Bx + C,$ the xcoordinate of the vertex is $x = B/2A.$
When we simply add or subtract a constant to/from the end of a function (the k in the vertex form in the box above), the effect on the graph is to raise or lower the whole graph by k units.
Manipulate the slider on the graph to change k in
$$f(x) = x^2 + k$$
If k is negative, the translation is downward, and if k is positive the translation is upward.
Notice that $f(x) = x^2 + k$ is just $f(x) = x^2,$ our parent function, with $k$ added to every $y$ value it produces. The vertical translation is probably the easiest transition to understand, and it works exactly the same way in any function.
When we subtract a constant, $h,$ from the independent variable of any function $(x  h),$ the effect is to translate it left or right (See the functions page to brush up on this). If $h \lt 0,$ the translation is to the left, and if $h \gt 0$ it's to the right. Note that we subtract $h$ from $x$ before squaring (PEMDAS!).
Recall that this "feels" backward: $(x  2)^2$ is a $x^2$ shifted two units to the right, and $(x + 2)^2$ is a $x^2$ shifted two units to the left, because it's really $(x  (2))^2.$ Notice that on when $x$ is negative, we're subtracting a negative number inside the parenthesis: $(x  h) = (x + h).$
Move the slider around on the graph to see the effect of changing $h.$ Notice what happens when $h$ gets bigger in both the positive and negative directions.
In this example, horizontal and vertical translations of 2 units in the positive direction are combined. We can do this sort of thing in exactly the same way for any function at all. More things are the same in mathematics than different!
The function looks like
When we multiply a function by a constant, A, the effect is to scale (expand or shrink) the graph vertically. If A > 1, the function is stretched vertically. If A < 1 it is compressed vertically, and if A is negative, it still scales the graph by A, but it is also flipped across the xaxis. For example, f(x) = x^{2} has a vertex of (0, 0) and opens upward, but g(x) = x^{2} opens downward – but has the same vertex.
Move the slider back and forth to change the value of the A parameter and watch its effect on the function.
When we divide the independent variable of a function by a constant number, c, the effect on its graph is horizontal scaling (stretching or compressing). For example, if x is divided by a number greater than 1, the graph is stretched horizontally. If it's divided by a number less than 1 (like ¼ in the plot here) the graph is compressed horizontally.
Move the slider back and forth to see the effect of the parameter c on the function. Notice that horizontal stretching looks a lot like vertical compression, and horizontal compression looks a lot like vertical stretching.
As with all functions, all of the various transformations can be combined for quadratic functions. Starting with the parent function, we can form any quadratic like this:
These transformations apply to any kind of function, and they always do the same things. Learn them now so that you'll easily be able to apply and interpret them with the other functions you'll learn along the way.
We have to be careful about the effect of horizontal scaling when we look for the vertex of a quadratic function.
Here's a quadratic that looks like it's stretched by a factor of two in the horizontal direction (accomplished by dividing the independent variable (x) by 2 where it occurs inside the function), and has a vertex of (2, 3).
$$f(x) = \left( \frac{x}{2}  2 \right)^2 + 3$$
It looks like the vertex of this function should be (2, 3), but it's not. If we expand this function to standard $(f(x) = Ax^2 + Bx + C$ form, we get
$$f(x) = \frac{1}{4} x^2  2x + 7$$
Now we can determine the xcoordinate of the vertex using x = B/2A, and it's not 2:
$$\frac{B}{2A} = \frac{1}{2 \cdot \frac{1}{4}} = 4$$
Plugging 4 into the function gives us the ycoordinate,
$$f(4) = \frac{16}{4}  8 + 7 = 3$$
which is what we'd expect, y = 3. The vertex of this parabola is at (4, 3). So what's going on? That horizontal stretching transformation has moved the vertex horizontally, too. Here's why:
When we multiply the binomial that contains the transformation, that divisor of 2 also contributes to the B term of the function, and therefore affects the xcoordinate of the vertex. Those mixed terms can move the vertex. It's that 2x that's moved the vertex from what we'd expected.
$$= \frac{x^2}{4}  \bf{2x} + 4$$
You'll need to become very good at quickly sketching quadratic functions. There are many clues about what the graph of a quadratic looks like; here's a first example.
This moves the vertex of the parabola from (0, 0) to (3, 1). Notice that when a quadratic function is in vertex form, we can easily just "read" the vertex from the function definition. The vertical scaling of 2 has the effect of stretching the parabola upward (by a factor of 2) from the vertex.
Now it's easy to sketch the graph →
It's often very useful to sketch the parent function onto a graph first, then follow the transformations to sketch the new function.
The xcoordinate of the vertex is:
The we obtain the ycoordinate simply by plugging x = 1 into the function:
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Now because A > 0, the parabola opens upward. It's also easy to find the yintercept, the point where x = 0: f(0) = 1.
We'll do more graphing of quadratics below, but first we'll figure out how to solve quadratic equations.
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We are often faced with finding the solution to the equation f(x) = k, where k is some constant. For example, we might want to solve the quadratic equation
3x^{2}  4x + 2 = 19,
But an equation like this can always be rewritten by
moving all of the constants to the left side:
3x^{2}  4x  17 = 0.
Now we're just finding the xintercepts, which can be done by direct factoring (we do this "by eye" from experience at recognizing patterns) or by completing the square. We'll develop both of these techniques below.
We think of direct factoring as finding binomial factors of a quadratic function "by eye" or "by instinct," or by the guessandcheck method. Whatever you call it, if it's possible, direct factoring is a quick way to the roots of a quadratic function.
Consider the factored quadratic function
f(x) = (x + m)(x + n)
= x^{2} + mx + nx + mn
= x^{2} + (m + n)x + mn,
which looks like Ax^{2} + Bx + C, with A = 1. Notice the correspondence: A = 1, B = m+n, and C = mn.
From this relation, you can easily see that what we're looking for in direct factoring is two numbers, m and n, that add to B and multiply to C.
The trick is to get good at recognizing patterns that lead us to proper factoring. That comes through practice  although on rare occasions, I've met people who are factoring savants.
Direct factoring when A ≠ 1 is a little trickier, but you can still learn to recognize the patterns:
f(x) = (ax + m)(bx +n)
= ab x^{2} + (mb + na)x + mn.
Here is a complete picture of the correspondence between the parameters A, B & C and a, b, m & n:
Now a·b must multiply to A, bm+an must sum to B and mn must multiply to C. Honestly, though, I never think of it this way; I factor this kind of quadratic equation by "educated trial and error."
Here's an interesting note on direct factoring. I once made a spreadsheet of quadratic functions with coefficients A= 9, B= 9, C= 9; A= 9, B =9, C= 8; ... up to A= 9, B= 9, C= 9. Of over 5000 possible quadratic equations that were generated in this way, only about 12% were directly factorable. As you work out in the world, not only is it unlikely that you'll encounter a factorable quadratic equation, but it's even less likely you'll encounter one with nice integer coefficients!
So of what value is direct factoring? Well, it's always good to be able to recognize patterns in mathematics, and factoring practice helps with that. It also helps us to arrive rapidly at exact solutions for many equations. But we'll want a more general way of solving quadratic equations, and that's completing the square.
The most general method of finding the solutions to Ax^{2} + Bx + C = 0 is the method called completing the square. It works like this: Take any binomial and square it, (x + m)^{2}, to form a perfect square. Then expand it:
(x + m)^{2} = x^{2} + 2mx + m^{2}.
Now notice that for a perfect square, the constant term is equal to the square of onehalf of the coefficient of x: m^{2} = [(1/2)(2m)]^{2}.
We can use this to our advantage. We can build a perfect square out of any quadratic function, but at the expense of accumulating some constants on the right side of the equation. But that's OK, we can easily deal with that later. Below are a few stepbystep examples of completing the square. We use the same basic steps each time, and you should learn them.
Completing the square is basically forcing part of a quadratic equation (the part with the variables) to be a perfect square with a well known pattern. The cost of this is accumulating constants on the other side of the equation, but we can always deal with the at the end.
We'll do these problems using the same steps each time. The first is
We begin with our function, and move the constant to the right side, in this case by addition of 35 to each side:
We divide by the coefficient of x^{2} to make it 1. In this case, it's already 1, so we move on ...
This is the key step. One half of 2 is 1, so we'll add 1 to both sides. I like to write the number added to the left as 1^{2} (of course when it's 1 it doesn't make much difference, but it will later), and just go ahead and square it on the right. Now we've made a perfect square on the left.
Now we've constructed a perfect square on the left. That's what we set out to do. The "cost" was that we had to add that 1 to the right side. Now our equation is
The green items are a consistent pattern for how to identify the perfect square in all of these problems: Take the x, the first operation (+ in this case) and the number that's 1/2 of the coefficient of x – before it's squared. The perfectsquare equation is:
Now it's easy to solve for x, simply take the square root of each side, remembering to append a ± to the right side (it could go on either side, but there works best).
Now it's a simple matter to find the two solutions, and you can see that they match our original solutions that came from factoring.
We begin with our function and walk through the steps:
We divide by the coefficient of x^{2} , 2 in this case:
One half of 1 is ½, so we'll add the square of ½ to both sides, again writing it as (½)^{2} on the left, and ¼ on the right. Doing so really helps to properly identify the perfect square on the left in the next step.
The perfect square is composed of the x from x^{2}, the sign following it ( + ) and the ½ from (½)^{2}:
Now it's easy to solve for x. Take the square root of each side, remembering to append a ± to the right side:
Now we just move that ½ to the right side by subtraction to get our two solutions:.
It is very unlikely that anyone would come up with these solutions by direct factoring!
If you haven't studied complex numbers yet, you might want to review them here.
We divide by the coefficient of x^{2} , 2 in this case:
One half of 5/2 is 5/4, so we'll add the square of 5/4 to both sides, again writing it as (5/4)^{2} on the left, and 25/16 on the right. Doing so really helps to properly identify the perfect square on the left in the next step.
The perfect square is composed of the x from x^{2}, the sign following it ( + ) and the 5/4 from (5/4)^{2}. We can also add the fractions on the right by using 16 as a common denominator:
Take the square root of each side, remembering to append a ± to the right side. Note that we're taking the square root of a negative number, so we'll need imaginary numbers:
Taking the root and moving the 5/4 to the right by subtraction gives a pair of imaginary roots:
Finally, let's complete the square on a completely general (no numbers) quadratic equation, Ax^{2} + Bx + C = 0. You'll see why this is relevant when we're done.
One half of B/A is B/(2A), (and remember that the square of that is B^{2}/(4A^{2}) so we get:
Here is the perfect square, and I've multiplied the C/A term on the right by 4A/4A to get a common denominator to add those terms:
Take the square root of each side, remembering to append a ± to the right side.
Now the 4A^{2} on the right side denominator is a square, so we can take its square root. Continuing the rearrangement by subtracting B/2A from both sides we arrive at:
which is the wellknown quadratic formula. Yup, completing the square and using the quadratic formula are the same thing!
The quadratic formula gives the two zeros or roots of any quadratic equation,
Ax^{2} + Bx + C = 0. It is:
Why do I need to be able to complete the square when I can just use the quadratic formula? Over the years I've seen many, many mistakes made using the quadratic formula. I think it has to do with my students overestimating how easy it is to use. But in fact, it's a little complicated, and all of the usual rules of algebra apply. For my part, I can complete the square as fast as anyone can use the quadratic, and I like the practice it gives me at adding fractions.
There are also situations you'll encounter later on, like when you study circles, ellipses and hyperbolas, where completing the square will help you immensely.
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It's easy to plot the graph of a quadratic function on a calculator or computer, but being able to do that doesn't guarantee that you'd notice if you keyed something in incorrectly. You will go far if you already have some notion (the more complete the better) of what a function should look like before you ever plot it on a machine.
Pro tip: Anybody can punch a function into a calculator or computer and get a pretty graph. Not just anyone can be sure it's the right graph. Be that person.
To sketch the graph of a quadratic function, you only need to do three things:
1. Find the vertex. First the xcoordinate using x = b/2a, then the y, using y = f(x).
2. Find the yintercept, f(0). At this point, you can make a decision. If the yintercept is above a vertex that is itself above the xaxis, then there is no way your parabola can ever cross the xaxis, therefore it must have complex roots, and there's no point in finding them  they won't help you sketch the graph.
3. If the opposite is true, you need to find the roots in order to mark the xaxis crossings. That can be done by completing the square or using the quadratic formula, or  if you're lucky  by direct factoring.
Below are a couple of examples; one is a function with complex roots. You should practice enough to get good at finding these points and sketching approximate graphs of quadratic functions.
There is a way to know fairly quickly whether the roots of a quadratic function are real: by calculating the discriminant. The discriminant is just the square root part of the quadratic formula, B^{2}  4AC →.
We have to resort to imaginary numbers if the discriminant is negative, and if it's positive we're assured of real roots. Here are all of the options:
If B^{2}  4AC > 0, the function has two real roots.
If B^{2}  4AC = 0, there is one double root, and the vertex of the parabola touches the xaxis.
If B^{2}  4AC < 0, there are two complex roots, a complexconjugate pair.
If a function has a double root, it touches the xaxis at only one point, its vertex. The same is true for any polynomial function with a double root: It will only touch the axis (not cross) at that point.
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Through any three points, only one unique parabola can be drawn. That may be a bit hard to come to terms with at first, but remember that a parabola isn't just any curve. It's a very specific kind of curve, the graph of a quadratic function. We'd like to be able to find the equation of that parabola.
Let's say that a parabola passes through three known points, (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}). Now we know that the general form of the equation we're looking for is
If each of out three points satisfies this equation for a certain set of A, B and C (the things we're really looking for), we have three equations:
Now that's three equations and three unknowns (A, B, C), so we should have enough information to determine what they are.
It's probably best at this point to work an example or two, so let's find the equation of the parabola above, which passes through
Here are our three equations:
These three equations can be solved simultaneously by methods you already know. We first use two sets of two equations to eliminate one of the variables (C is usually easiest in these problems), then use those two equations to solve for the other two, and finally use the results to find the third variable.
Here's how it looks. If we solve all three equations for C, we get:
Now set two pairs of equations equal to one another, eliminating C: equations 1 and 2, 2 and 3:
and rearrange to:
Now we've reduced the problem to a twoequations, twounknowns one. Let's multiply the lower equation by 5 to get
If we add those (remember that the left sides are equal to the right sides, so that has to be true for the sums, too), we get
35A = 17, or A = 17/35.
Then it's easy to plug in A to get B = 33/35 and C = 124/35. So the equation of the parabola that fits through our three points is
f(x) = (17/35)x^{2}  (33/35)x  124/35
which matches the graph above.
Minutes of your life: 2:42
Minutes of your life: 2:50
In this example we complete the square to find the roots of a function for which we already know the roots. It's a good example to work through to see that completing the square works!
Minutes of your life: 2:46
In this example, the quadratic function f(x) = 8x^{2}  20x  3 has real roots, but it's not factorable. We complete the square to find the exact roots of the function. There is no difference between completing the square and using the quadratic formula, but I think that knowing how to complete the square will help you to understand a lot about basic algebra steps. You should learn it!
Minutes of your life: 2:58
In this example, we start with f(x) = Ax^{2} + Bx + C and complete the square to derive the quadratic formula.
Minutes of your life: 2:04 (and well worth it!)
In this example, we find the roots of f(x) = 3x^{2}  4x + 7, which are imaginary (or more accurately, are complex with nonzero imaginary parts. We also use the discriminant to show how to determine beforehand that the roots of a quadratic function will be nonreal.
Minutes of your life: 2:51
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