Drag the vertex
Completing the square: standard form to vertex form
- 1. Start from the functiony = x² − 2x − 3
Controls
Form of the equation
Readings
- Vertex (h, k)
- (1, −4)
- Axis of symmetry
- x = 1
- Discriminant, D = b² − 4ac
- 16
- Nature of roots
- Two different real roots
- Roots by the quadratic formula
- x = (2 ± √16) / 2 = 3, −1
- y-intercept
- (0, −3)
- Parabola opens
- Upwards (a > 0)
- Sum of roots, −b/a
- 2
- Product of roots, c/a
- −3
- Moving point (x, y)
- (0, −3)
How to use this simulation
- Start by just watching: the screen shows y = x² − 2x − 3. The parabola crosses the x-axis at −1 and 3, and those two green dots are the roots.
- Drag the a slider below zero and the parabola flips to open downwards. Set a to exactly 0 and the curve straightens into a line.
- Raise c to lift the whole curve, and watch the discriminant D in the readings: D = 0 at the moment it touches the x-axis, and negative once it floats above it.
- Grab the orange vertex with your finger or mouse and move it, or pick “Vertex form” and use the h and k sliders. The axis of symmetry follows.
- Keep “Show completing the square” on: five steps appear one by one below the graph, using your own numbers, and end in vertex form.
- Pick “Factored form”, set the roots α and β directly, and check the sum −b/a and product c/a in the readings.
Throw a ball, watch a fountain: the same curve everywhere
Toss a ball up and a little forward. It slows as it rises, pauses for an instant at the top, and falls back down. Trace its path and you get a smooth upside-down U. Mathematicians call that shape a parabola, and the equation behind it is a quadratic.
The arc of water from a drinking fountain, the dish of a satellite antenna, the mirror inside a torch and the curve of some bridges all belong to the same family. Each one follows a rule of the form y = ax² + bx + c.
Shopkeepers use this maths without knowing it. Price something too low and the profit is thin; price it too high and nobody buys. Somewhere in the middle is the price that earns the most. A graph of profit against price is a downward parabola, and the best price is its vertex.
On this page you will change a, b and c in the simulation and see for yourself how the parabola moves, when it cuts the x-axis twice, when it just touches, and when it misses completely. Then you will learn three ways to solve a quadratic equation and work through a set of exam-style problems.
Starting from zero: what is a quadratic?
The degree of a term is how many times the variable is multiplied by itself: x has degree 1, x² has degree 2, x³ has degree 3. An equation whose highest power of the variable is 2 is called a quadratic equation. “Quadratic” comes from the Latin quadratus, a square, because the key term is x squared.
The standard form of a quadratic equation in one variable is ax² + bx + c = 0. Here a is the coefficient of x², b is the coefficient of x and c is the constant term. They are real numbers, and one condition is essential: a ≠ 0. If a were 0 the x² term would vanish and you would be left with bx + c = 0, which is a linear equation.
b or c can be zero without any problem. x² − 9 = 0 (where b = 0) and x² − 4x = 0 (where c = 0) are both quadratics. Only a is not allowed to be zero. Slide a to 0 in the simulation and the parabola straightens into a line, with a label saying it is no longer quadratic.
Keep the difference between an equation and a function clear. ax² + bx + c = 0 is an equation: it asks “which values of x make the left-hand side zero?” y = ax² + bx + c is a function: it gives one y for every x. The graph of the function is a parabola, and the x-values where it meets the x-axis (where y = 0) are exactly the solutions of the equation.
A root, zero or solution is a number that makes the equation true when you put it in place of x. A quadratic equation has at most two roots. Sometimes there are two different roots, sometimes the two roots are equal, and sometimes there are no real roots at all. The discriminant, coming up shortly, tells you which.
Key terms
Definition questions come straight from this list, so it is worth learning the words precisely.
| Term | Symbol | Meaning |
|---|---|---|
| Quadratic equation | ax² + bx + c = 0 | Highest power of the variable is 2, and a ≠ 0 |
| Coefficients | a, b | The numbers multiplying x² and x |
| Constant term | c | The term with no variable in it |
| Roots (zeros, solutions) | α, β | The values that make the equation true |
| Discriminant | D = b² − 4ac | Decides how many real roots there are |
| Parabola | — | The graph of y = ax² + bx + c, a U or an upside-down U |
| Vertex | (h, k) | The lowest or highest point of the parabola |
| Axis of symmetry | x = h | The vertical line through the vertex that splits the parabola into mirror halves |
| Completing the square | (x + p)² | Rewriting the expression around a perfect square |
| y-intercept | (0, c) | Where the graph crosses the y-axis, found by putting x = 0 |
Method 1: solving by factoring
Factoring is usually the fastest method when it works. It rests on one simple fact, the zero product property: if two numbers multiply to zero, at least one of them must be zero. 3 × 0 = 0 and 0 × 7 = 0, but 3 × 7 is never 0.
So if you can write ax² + bx + c as a product of two linear factors, such as (x − p)(x − q) = 0, then either x − p = 0 or x − q = 0. That gives x = p or x = q. One quadratic has become two tiny linear equations.
When a = 1, look for two numbers that multiply to c and add to b, then split the middle term bx using them. When a ≠ 1, the two numbers must multiply to a × c and add to b (the “ac method”), and you group the four terms in pairs.
Factoring has limits. If the roots are awkward fractions or irrational numbers such as 2 + √3, you will never find neat integer factors. That is when completing the square or the formula takes over. It is also why the simulation’s factored form only works when there are real roots.
ax² + bx + c = a(x − α)(x − β)α and β are the roots
pq = 0 ⇒ p = 0 or q = 0zero product property
Method 2: completing the square
Completing the square means rearranging the equation so that one side is a perfect square (x + p)². Why bother? Because once you have (x + p)² = r, you take square roots to get x + p = ±√r, and x is one step away.
The trick comes from the identity (x + p)² = x² + 2px + p². The coefficient of x is 2p and the last term is p². So if you take half the coefficient of x and square it, adding that number turns the expression into a perfect square. x² + 6x plus (6/2)² = 9 gives x² + 6x + 9 = (x + 3)².
The steps are always the same: (1) divide the whole equation by a so that x² has coefficient 1; (2) move the constant to the right-hand side; (3) add (half the x coefficient)² to both sides; (4) write the left side as (x + p)²; (5) take square roots, remembering the ± sign.
This method works for every quadratic, whether or not it factors. Even more important, it is where the quadratic formula comes from: run exactly these steps on the general equation ax² + bx + c = 0 and the formula drops out. Below the simulation the same steps run on whatever numbers you choose.
Method 3: the quadratic formula, derived
Below, completing the square is applied to the general equation ax² + bx + c = 0. Each line follows from the one before, so read it as an argument rather than something to memorise. (In India this is often called Sridharacharya’s formula, after the mathematician who described the method more than a thousand years ago.)
In the first step we divide by a, which is allowed because a ≠ 0. Next c/a moves to the right. Then (b/2a)² is added to both sides, since half of the x coefficient b/a is b/2a. Now the left side is a perfect square, and the right side is put over the common denominator 4a².
Taking the square root of both sides brings in ±, because every positive number has two square roots (the square roots of 9 are +3 and −3). Moving b/2a across finishes the job. Notice what is under the root: b² − 4ac, the discriminant. If it is negative the square root is not a real number, so there are no real roots.
ax² + bx + c = 0a ≠ 0
x² + (b/a)x + c/a = 0divide by a
x² + (b/a)x = −c/aconstant to the right
x² + (b/a)x + (b/2a)² = (b/2a)² − c/aadd (b/2a)² to both sides
(x + b/2a)² = (b² − 4ac) / 4a²perfect square
x + b/2a = ±√(b² − 4ac) / 2asquare root, with ±
x = (−b ± √(b² − 4ac)) / 2athe quadratic formula
The discriminant: the nature of the roots without solving
The expression under the square root, b² − 4ac, is called the discriminant and written D (or Δ). It discriminates between the possible kinds of answer: from its sign alone you know how many real roots there are and what they are like, before you solve anything.
If D > 0, √D is a positive real number, so the + and − give two different real roots, and the parabola cuts the x-axis at two points. If D is also a perfect square (16, 25, …) and a, b, c are rational, the roots are rational and the quadratic factors nicely. If not, the roots are irrational.
If D = 0, √D = 0 and both signs give the same answer, x = −b/2a. We say there are two equal real roots (a repeated root). On the graph the parabola touches the x-axis at its vertex without crossing it.
If D < 0, you would need the square root of a negative number, which is not real. So there are no real roots and the parabola never meets the x-axis. Later, once you meet complex numbers, you will say the roots form a complex conjugate pair.
| Discriminant | Nature of roots | What the graph does |
|---|---|---|
| D > 0, a perfect square | Real, different and rational | Crosses the x-axis twice |
| D > 0, not a perfect square | Real, different and irrational | Crosses the x-axis twice |
| D = 0 | Real and equal | Touches the x-axis at the vertex |
| D < 0 | No real roots (complex pair) | Never meets the x-axis |
Sum and product of the roots
If α and β are the roots of ax² + bx + c = 0, adding the two formula answers cancels the ± parts and leaves α + β = −b/a. Multiplying them uses (p + q)(p − q) = p² − q² and gives αβ = c/a. These two relations (sometimes called Vieta’s formulas) appear constantly in exams.
Their power is that you can answer questions about the roots without finding them. For example α² + β² = (α + β)² − 2αβ, and 1/α + 1/β = (α + β)/αβ. With irrational roots this saves a lot of messy arithmetic.
They also work in reverse. If you know the roots, the equation is x² − (sum of roots)x + (product of roots) = 0, because multiplying out (x − α)(x − β) gives exactly x² − (α + β)x + αβ.
α + β = −b/asum of roots
αβ = c/aproduct of roots
x² − (α + β)x + αβ = 0equation from its roots
The graph: parabola, vertex and axis of symmetry
The graph of y = ax² + bx + c is always a parabola. If a > 0 it opens upwards like a U, the vertex is the lowest point and the function has a minimum value. If a < 0 it opens downwards, the vertex is the highest point and the function has a maximum value.
The bigger |a| is, the narrower and steeper the parabola; the smaller |a| is, the wider and flatter. In the simulation, bring a down from 3 to 0.5 and watch the curve spread out.
The vertex is (h, k) with h = −b/2a and k = c − b²/4a (or just substitute x = h to find y). The vertical line x = h through the vertex is the axis of symmetry: fold the graph along it and the two arms match exactly. So when there are two roots they sit the same distance either side of the axis, and h is their average.
Writing the function as y = a(x − h)² + k, found by completing the square, is called vertex form. Because (x − h)² is never negative, when a > 0 the smallest value of y happens at x = h and equals k. It is the quickest form for reading the vertex at a glance.
The effect of b is the surprising one: changing b moves the vertex along a curved path rather than sliding the whole parabola sideways. And c is always the y-intercept, because putting x = 0 leaves y = c. On the starting graph that is (0, −3).
h = −b / 2a, k = c − b² / 4avertex (h, k)
y = a(x − h)² + kvertex form
x = haxis of symmetry
| x | y = x² − 2x − 3 | distance from the axis, x − h |
|---|---|---|
| −2 | 5 | −3 |
| −1 | 0 | −2 |
| 0 | −3 | −1 |
| 1 | −4 | 0 |
| 2 | −3 | 1 |
| 3 | 0 | 2 |
| 4 | 5 | 3 |
Try this in the simulation
Each experiment takes a minute or two. Predict what will happen first, then check.
Experiment 1: slide the parabola up and down with c
Stay in standard form and increase only c. The whole parabola rises without changing shape. The two roots move towards each other, merge into one (D = 0) and then disappear (D < 0).
At which c does it just touch? From b² − 4ac = 0, c = b²/4a. Work it out for the starting a and b, then check with the slider.
Experiment 2: narrow, wide and upside down with a
Lower a slowly from 3. The parabola widens, becomes almost flat near 0, is a straight line at exactly 0, and opens downwards once a is negative.
At a = 0 the vertex and the axis of symmetry disappear, because a line has no vertex. There is then only one root, x = −c/b.
Experiment 3: drag the vertex
With a > 0, grab the orange vertex and lift it above the x-axis. The roots merge and vanish, and the readings say there are no real roots. Pull it back down and they return.
Watch b and c change as you drag: from vertex form, b = −2ah and c = ah² + k.
Experiment 4: build an equation from its roots
Choose “Factored form” and set α = 2, β = 5. With a = 1 the equation is x² − 7x + 10 = 0. Check that the sum reads 7 and the product 10.
Now set a = 2. The roots stay put but the parabola gets narrower: infinitely many quadratics share the same roots and differ only in a.
Experiment 5: see the symmetry with the moving point
Turn on the moving point and watch (x, y) in the readings. When the point is the same distance either side of the axis, y is the same. As it passes the vertex, y turns from falling to rising (or the other way round).
Worked examples (every number computed)
Each problem is laid out the way an examiner likes to see it. The first one is the simulation’s starting graph, so you can compare with the readings on screen.
Example 1: everything about y = x² − 2x − 3
Here a = 1, b = −2, c = −3. The discriminant is D = b² − 4ac = 4 − (−12) = 16 > 0 and a perfect square, so the roots are real, different and rational.
By the formula x = (2 ± √16) / 2 = (2 ± 4) / 2, so x = 3 or x = −1. In factored form y = (x + 1)(x − 3).
Vertex: h = −b/2a = 1, k = −4; axis of symmetry x = 1; a > 0 so it opens upwards with minimum value −4. y-intercept (0, −3). Sum −b/a = 2, product c/a = −3.
Example 2: solve x² − 5x + 6 = 0 by factoring
We want two numbers that multiply to 6 and add to −5. They are −2 and −3, so split the middle term: x² − 2x − 3x + 6 = 0.
Grouping gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. Check: the sum and product of these roots match −b/a and c/a.
Example 3: solve 2x² + 8x − 10 = 0 by completing the square
Divide by 2: x² + 4x − 5 = 0, so x² + 4x = 5. Half the x coefficient is 2 and its square is 4; add it to both sides: x² + 4x + 4 = 5 + 4.
So (x + 2)² = 9, and taking square roots x + 2 = ±3. Hence x = 1 or x = −5.
Example 4: x² − 4x + 1 = 0 by the formula (irrational roots)
D = 12, not a perfect square, so the roots are irrational. x = (4 ± √12) / 2 = 2 ± √3.
As decimals √12 ≈ 3.46, so x ≈ 3.73 or x ≈ 0.27. This one does not factor over the integers, so the formula is the right tool.
Example 5: nature of roots and the condition for equal roots
The discriminant tells us the nature of the roots of these three equations without solving them:
2x² − 6x + 3 = 0: D = 36 − 24 = 12 > 0 and not a perfect square, so real, different, irrational roots. 4x² − 12x + 9 = 0: D = 144 − 144 = 0, so real, equal roots. x² + x + 1 = 0: D = 1 − 4 = −3 < 0, so no real roots.
For which k does x² + kx + 16 = 0 have equal roots? Set D = 0: k² − 64 = 0, so k = ±8.
Example 6: α² + β² and 1/α + 1/β for 3x² − 7x + 2 = 0
α + β = −b/a = 7/3 and αβ = c/a = 2/3.
α² + β² = (α + β)² − 2αβ = 49/9 − 12/9 = 37/9 ≈ 4.11. And 1/α + 1/β = (α + β)/αβ = 7/2 = 3.5.
Check: the formula gives roots 2 and about 0.333, and the sum of their squares comes out the same.
Example 7: the equation whose roots are 4 and −3
Sum = 1, product = −12. Using x² − (sum)x + product = 0, the equation is x² − x − 12 = 0.
Example 8: sides of a rectangle from its area
A rectangular garden is 3 m longer than it is wide, and its area is 40 m². Let the width be x: x(x + 3) = 40, so x² + 3x − 40 = 0.
D = 169, √D = 13, so x = (−3 ± 13) / 2 = 5 or −8. A length cannot be negative, so the width is 5 m and the length 8 m. Check: 5 × 8 = 40.
Example 9: the height of a ball thrown upwards
A ball is thrown straight up at 20 m/s. Taking g ≈ 10 m/s², its height after t seconds is h = 20t − 5t². When is it 15 m high?
Setting 20t − 5t² = 15 gives 5t² − 20t + 15 = 0; dividing by 5 gives t² − 4t + 3 = 0. So t = 1 s (on the way up) and t = 3 s (on the way down). Both answers make sense.
The highest point is the vertex: at t = 2 s, h = 20 m. And h = 0 at t = 0 and t = 4 s, so the ball is back in the hand after 4 seconds.
Example 10: the price that maximises profit
A stall’s daily profit is P = −2x² + 40x − 150 (in thousands) when the selling price is x. Since a < 0 the parabola opens downwards and the maximum profit is at the vertex.
x = −b/2a = 10 gives the maximum profit P = 50 thousand. Profit is zero (break-even) when x² − 20x + 75 = 0, that is at x = 5 or 15; any price outside that range makes a loss.
Example 11: write y = 2x² − 12x + 13 in vertex form
Take 2 out of the x terms: y = 2(x² − 6x) + 13. Half the x coefficient inside the bracket is −3 and its square is 9; add and subtract it and tidy up: y = 2(x − 3)² − 5.
So the vertex is (3, −5) and the minimum value is −5. D = 40 > 0, and the roots x ≈ 4.58 and 1.42 sit the same distance either side of x = 3.
Common mistakes
These cost marks every year. Knowing them in advance is half the battle.
- Getting the sign of −b wrong: if b = −2 then −b = +2, not −2.
- Forgetting that −4ac becomes positive when a and c have opposite signs.
- Taking only the positive square root and losing one of the two roots.
- Dividing only the square-root part by 2a instead of the whole numerator.
- Dividing both sides of x² = 5x by x, which throws away the root x = 0. Write x(x − 5) = 0 instead.
- Reading off a, b and c before rearranging the equation into ax² + bx + c = 0.
- Keeping a root that makes no sense in a word problem, such as a negative length or time.
- Writing the vertex x-coordinate as b/2a instead of −b/2a.
Quadratics in real life
Parabolas and quadratic equations are hiding all around you:
- The path of a thrown ball, a football kick or a jet of water is a parabola if you ignore air resistance.
- Satellite dishes and torch reflectors are parabolic because a parabola gathers parallel rays at a single point, the focus.
- Businesses find maximum profit or minimum cost at the vertex of a quadratic model.
- Builders and designers find side lengths from an area, which leads straight to a quadratic.
- A car’s braking distance grows with the square of its speed, so doubling your speed roughly quadruples the distance.
- The cables of a suspension bridge hang in a shape very close to a parabola.
Exam corner
Most exam systems test quadratics in three ways: solving (by factoring, completing the square or the formula), analysing (discriminant, vertex, axis, intercepts) and modelling (area, projectile and profit problems). Check which methods your syllabus expects you to show: some papers insist on completing the square for particular questions.
For graph sketching, collect the key features before drawing: the direction of opening from the sign of a, the y-intercept (0, c), the roots if D ≥ 0, and the vertex. Plot these, draw a smooth symmetric curve, and label everything.
For multiple-choice questions, remember the shortcuts: sum of roots −b/a, product c/a; D = 0 means equal roots; and if a and c have opposite signs then D is automatically positive, so there are certainly two real roots.
One-screen revision summary
Everything you need the night before the exam:
- Standard form: ax² + bx + c = 0, with a ≠ 0.
- Three methods: factoring, completing the square, the quadratic formula.
- Formula: x = (−b ± √(b² − 4ac)) / 2a.
- Discriminant D = b² − 4ac: > 0 two real roots, = 0 equal roots, < 0 no real roots.
- α + β = −b/a, αβ = c/a; the equation from its roots is x² − (α + β)x + αβ = 0.
- The graph is a parabola; a > 0 opens up, a < 0 opens down.
- Vertex (−b/2a, c − b²/4a); axis of symmetry x = −b/2a; y-intercept (0, c).
- Vertex form y = a(x − h)² + k.
Frequently asked questions
What is a quadratic function?
A function of the form y = ax² + bx + c where a, b and c are real numbers and a ≠ 0. Its graph is a parabola.
What is the quadratic formula?
x = (−b ± √(b² − 4ac)) / 2a. It solves every quadratic equation ax² + bx + c = 0 and is derived by completing the square.
What is the discriminant?
The expression b² − 4ac. If it is positive there are two real roots, if it is zero there is one repeated real root, and if it is negative there are no real roots.
How do you find the vertex of a parabola?
Work out h = −b/2a, then substitute x = h into the function to get k. The vertex is (h, k), and the axis of symmetry is the line x = h.
What is vertex form?
y = a(x − h)² + k, where (h, k) is the vertex. You reach it from standard form by completing the square.
What do the sum and product of the roots equal?
For ax² + bx + c = 0 with roots α and β, α + β = −b/a and αβ = c/a.
How do you write a quadratic from its roots?
Use x² − (sum of roots)x + (product of roots) = 0. For roots 2 and 3 that is x² − 5x + 6 = 0; multiply by any non-zero a for other equations with the same roots.
Why must a not be zero?
If a = 0 the x² term disappears, leaving the linear equation bx + c = 0, whose graph is a straight line rather than a parabola.
How does a change the graph?
The sign of a sets the direction: positive opens upwards, negative opens downwards. The size of a sets the width: a larger |a| makes a narrower parabola.
Which method should I use to solve a quadratic?
Try factoring first when the numbers are small and D is a perfect square. Otherwise use the formula. Use completing the square when you need vertex form or the question asks for it.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
Textbook
Class 9-10 Higher Mathematics textbook
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CUET admission 2026-27
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SSC Mathematics: syllabus and preparation
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SSC Higher Mathematics: syllabus and preparation
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HSC Higher Mathematics: syllabus and preparation
Textbook
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Engineering admission test preparation
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English
Paragraph: Artificial Intelligence
Bangla
রচনা: কৃত্রিম বুদ্ধিমত্তা (in Bangla)
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SSC suggestions
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HSC suggestions
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SSC syllabus
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HSC syllabus
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