A photo quiz
by AdamM7.
Estimated time: 5 mins.

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Quiz Answer Key and Fun Facts

Answer:
**Blue**

The blue line is steeper, so it has a faster rate of change, and thus a larger gradient.

A gradient of 2 tells you that for every 1 square you move to the right, you move 2 squares upwards. That is, the ratio of growth of y to x is 2. A gradient of 1, then, is a line which moves rightwards at the same speed it moves upwards - a line at a 45° angle from the horizontal.

We can see that every time you move 1 square to the right on the blue line, you have to move 2 squares upwards, so the blue line has a gradient of 2. Similarly, the red line moves 1 upwards for every 1 you move to the right, so its gradient is 1. So we can also reach the answer by calculating gradients and comparing (as 2 is greater than 1).

The blue line is steeper, so it has a faster rate of change, and thus a larger gradient.

A gradient of 2 tells you that for every 1 square you move to the right, you move 2 squares upwards. That is, the ratio of growth of y to x is 2. A gradient of 1, then, is a line which moves rightwards at the same speed it moves upwards - a line at a 45° angle from the horizontal.

We can see that every time you move 1 square to the right on the blue line, you have to move 2 squares upwards, so the blue line has a gradient of 2. Similarly, the red line moves 1 upwards for every 1 you move to the right, so its gradient is 1. So we can also reach the answer by calculating gradients and comparing (as 2 is greater than 1).

Answer:
**Negative**

The line is decreasing, so the gradient is negative.

If the gradient was positive, the line would be going up. If it was zero, the line can't be going up or down (zero is neither positive nor negative), so it must be horizontal. A gradient of infinity corresponds to a vertical line.

We can check that these ideas accord with the formal definition of gradients as the ratio of the change in y to the change in x. For the red line, every 1 you move to the right, you move 1 downwards. To calculate the gradient we need to phrase this slightly unnaturally as "for every 1 you move to the right, you move -1 upwards", and then the ratio is -1. So the gradient is -1.

For the horizontal line, every 1 you move to the right corresponds to moving 0 upwards, so the gradient is 0. It doesn't grow or shrink at all. The vertical line, in contrast, grows infinitely fast: for every 1 square you move upwards, you move 0 squares to the right! If you were to calculate the ratio 1/0, you would get a math error: dividing by 0 does not make sense. The line is steeper than any line with a positive gradient, so we call its gradient "(positive) infinity".

The line is decreasing, so the gradient is negative.

If the gradient was positive, the line would be going up. If it was zero, the line can't be going up or down (zero is neither positive nor negative), so it must be horizontal. A gradient of infinity corresponds to a vertical line.

We can check that these ideas accord with the formal definition of gradients as the ratio of the change in y to the change in x. For the red line, every 1 you move to the right, you move 1 downwards. To calculate the gradient we need to phrase this slightly unnaturally as "for every 1 you move to the right, you move -1 upwards", and then the ratio is -1. So the gradient is -1.

For the horizontal line, every 1 you move to the right corresponds to moving 0 upwards, so the gradient is 0. It doesn't grow or shrink at all. The vertical line, in contrast, grows infinitely fast: for every 1 square you move upwards, you move 0 squares to the right! If you were to calculate the ratio 1/0, you would get a math error: dividing by 0 does not make sense. The line is steeper than any line with a positive gradient, so we call its gradient "(positive) infinity".

Answer:
**Blue (C)**

The curve is much steeper at the blue point - between the x-values -2 and -1, the curve just dips slightly below 0 and rises again, whereas from 0 to 1 it grows too large to be shown on the graph, above the y-value 5. This means the graph will have a much larger gradient at the blue point.

You might notice that the curve is symmetric: it has a line of symmetry at (the vertical line) x = -1. Because of this symmetry, A and C have the same height above the x-axis (i.e. the same y co-ordinates). Additionally, their gradients are negatives of each other: the curve has gradient -21/2 at A, and gradient 21/2 at C. This is because the curve must be moving downwards at the same speed that it later moves upwards at.

To those with existing calculus knowledge, the equation of this curve is y = x(x+2)(x+1)^2. The points are A = (-5/2, 45/16), B = (-3/2, -3/16) and C = (1/2, 45/16). The gradient of the curve has equation y = 4x^3 + 12x^2 + 10x + 2, from which we calculate that the gradient at A is -21/2, the gradient at B is 1/2 and the gradient at C is 21/2. This means the gradient is 21 times larger at C than at B.

The curve is much steeper at the blue point - between the x-values -2 and -1, the curve just dips slightly below 0 and rises again, whereas from 0 to 1 it grows too large to be shown on the graph, above the y-value 5. This means the graph will have a much larger gradient at the blue point.

You might notice that the curve is symmetric: it has a line of symmetry at (the vertical line) x = -1. Because of this symmetry, A and C have the same height above the x-axis (i.e. the same y co-ordinates). Additionally, their gradients are negatives of each other: the curve has gradient -21/2 at A, and gradient 21/2 at C. This is because the curve must be moving downwards at the same speed that it later moves upwards at.

To those with existing calculus knowledge, the equation of this curve is y = x(x+2)(x+1)^2. The points are A = (-5/2, 45/16), B = (-3/2, -3/16) and C = (1/2, 45/16). The gradient of the curve has equation y = 4x^3 + 12x^2 + 10x + 2, from which we calculate that the gradient at A is -21/2, the gradient at B is 1/2 and the gradient at C is 21/2. This means the gradient is 21 times larger at C than at B.

Answer:
**Red**

The blue and orange lines do not touch the point A at all, while the pink line touches an immediately surrounding point. The red line does touch the green curve again, but not at a point in the same "hump" that A is at the peak of, so this is not "immediately surrounding".

The tangent is important in calculus because it has the same gradient as the curve at the point. Remember from question 2 that a flat line has gradient 0, so the red tangent has a gradient of 0 and the point A on the green curve has gradient 0. These special points are known as "turning points", and it should be clear from looking at the graph why that is. In this example, the curve "turns around" and starts decreasing rather than increasing.

There are two other types of turning points. The first is when the curve turns around the other way, and goes from increasing to decreasing. The second is an "inflection point", where the curve motions as if it is about to turn around... and then starts growing again in the same direction. A web search will yield some examples if you are interested.

The blue and orange lines do not touch the point A at all, while the pink line touches an immediately surrounding point. The red line does touch the green curve again, but not at a point in the same "hump" that A is at the peak of, so this is not "immediately surrounding".

The tangent is important in calculus because it has the same gradient as the curve at the point. Remember from question 2 that a flat line has gradient 0, so the red tangent has a gradient of 0 and the point A on the green curve has gradient 0. These special points are known as "turning points", and it should be clear from looking at the graph why that is. In this example, the curve "turns around" and starts decreasing rather than increasing.

There are two other types of turning points. The first is when the curve turns around the other way, and goes from increasing to decreasing. The second is an "inflection point", where the curve motions as if it is about to turn around... and then starts growing again in the same direction. A web search will yield some examples if you are interested.

Answer:
**No **

The technical term for moving a shape or geometric object is "translating".

Translating a line does not change how steep it is - though rotating would - so translating does not change the gradient. In the given graph, we can see three lines of equal steepness (gradient 1/2, in fact) that are translations of one another.

For those who already know how to calculate the derivative of a function, notice that translating a graph only changes the constant term: the depicted lines are y = x/2+1, y = x/2 and y = x/2-1. The constant term disappears in differentiation, so it does not contribute towards the gradient value. This is because the gradient is a measure of rate of change.

The technical term for moving a shape or geometric object is "translating".

Translating a line does not change how steep it is - though rotating would - so translating does not change the gradient. In the given graph, we can see three lines of equal steepness (gradient 1/2, in fact) that are translations of one another.

For those who already know how to calculate the derivative of a function, notice that translating a graph only changes the constant term: the depicted lines are y = x/2+1, y = x/2 and y = x/2-1. The constant term disappears in differentiation, so it does not contribute towards the gradient value. This is because the gradient is a measure of rate of change.

Answer:
**5.7**

The red area is entirely above the x-axis, so we do not need to worry about subtracting areas or negative numbers. We can see 4 squares entirely in red, and then a small amount of area above those squares, and a more substantial amount of area to the right of those squares. Overall, it definitely covers more than 4 squares, and definitely less than 8 (as the red area only spills into 8 squares).

So, we can rule out all other possible answers as they are either more than 8 (8.4) or less than 4 (3.9 and -3.0). We do not even need to begin counting to rule out -3.0 as the area is positive.

"Counting squares" is a perfectly valid method of integration, though students are more likely to learn algebraic methods to find exact answers in particular cases. It is a sad reality, however, that many functions are too complicated for us to integrate algebraically, and so some sophisticated area-counting method (such as the trapezium rule) is the best we can do.

The red area is entirely above the x-axis, so we do not need to worry about subtracting areas or negative numbers. We can see 4 squares entirely in red, and then a small amount of area above those squares, and a more substantial amount of area to the right of those squares. Overall, it definitely covers more than 4 squares, and definitely less than 8 (as the red area only spills into 8 squares).

So, we can rule out all other possible answers as they are either more than 8 (8.4) or less than 4 (3.9 and -3.0). We do not even need to begin counting to rule out -3.0 as the area is positive.

"Counting squares" is a perfectly valid method of integration, though students are more likely to learn algebraic methods to find exact answers in particular cases. It is a sad reality, however, that many functions are too complicated for us to integrate algebraically, and so some sophisticated area-counting method (such as the trapezium rule) is the best we can do.

Answer:
**0**

The area under the x-axis contributes a negative value, so we have to find the area above the x-axis and then subtract away the area below the x-axis. Because of the symmetries of the graph, these two areas are the same size and shape (if you were to rotate one of them by 180°), so when we subtract one away from the other, we are left with 0.

The areas added together would amount to about 1.90 squares.

The graph, for those with existing calculus expertise, is a scaling of cosine so the period is 3 rather than pi radians i.e. y = cos(pi*x/3).

The area under the x-axis contributes a negative value, so we have to find the area above the x-axis and then subtract away the area below the x-axis. Because of the symmetries of the graph, these two areas are the same size and shape (if you were to rotate one of them by 180°), so when we subtract one away from the other, we are left with 0.

The areas added together would amount to about 1.90 squares.

The graph, for those with existing calculus expertise, is a scaling of cosine so the period is 3 rather than pi radians i.e. y = cos(pi*x/3).

Answer:
**1**

The function has equation y = (x^2+x)/x, and there is a discontinuity at x = 0 because if you type (0^2+0)/0 in your calculator, you'll get a math error (division by 0 doesn't work).

A function (line or curve or shape) is continuous if it has no discontinuities, which seems fairly straightforward. What is less straightforward is how to define this formally, and indeed mathematicians had no answer to this for the first century of calculus' existence. They eventually settled on a rather complex definition beyond the scope of this quiz.

Suffice it to say that the definition of continuous formalizes the idea that the function must be defined everywhere (unlike here, where division by 0 prevents this) and there can be no sudden jumps in value (like we will see in the next question). If you can draw something without lifting your pen from the paper, and it does not loop back on itself, then it is continuous. Moreover, it is actually a special kind of continuous known as "uniformly continuous".

The function has equation y = (x^2+x)/x, and there is a discontinuity at x = 0 because if you type (0^2+0)/0 in your calculator, you'll get a math error (division by 0 doesn't work).

A function (line or curve or shape) is continuous if it has no discontinuities, which seems fairly straightforward. What is less straightforward is how to define this formally, and indeed mathematicians had no answer to this for the first century of calculus' existence. They eventually settled on a rather complex definition beyond the scope of this quiz.

Suffice it to say that the definition of continuous formalizes the idea that the function must be defined everywhere (unlike here, where division by 0 prevents this) and there can be no sudden jumps in value (like we will see in the next question). If you can draw something without lifting your pen from the paper, and it does not loop back on itself, then it is continuous. Moreover, it is actually a special kind of continuous known as "uniformly continuous".

Answer:
**There is no limit value.**

Remember that a function needs to get closer and closer to the limit value as you approach it in each direction. If you were confused about what it is because it needs to equal 3 (the value if you approach from the left) and equal 1 at the same time (the value approaching from the right), that's a good confusion to have! The resolution is that there is no limit value. It can't possibly exist, because no number is both equal to 3 and equal to 1.

The value 2 is the x co-ordinate at the discontinuity. Formally, we have said "the function has no limit as x approaches 2".

This function is a piecewise function, with equation y = 3 when x is less than 2, and equation y = 1 when x is greater than 2.

Remember that a function needs to get closer and closer to the limit value as you approach it in each direction. If you were confused about what it is because it needs to equal 3 (the value if you approach from the left) and equal 1 at the same time (the value approaching from the right), that's a good confusion to have! The resolution is that there is no limit value. It can't possibly exist, because no number is both equal to 3 and equal to 1.

The value 2 is the x co-ordinate at the discontinuity. Formally, we have said "the function has no limit as x approaches 2".

This function is a piecewise function, with equation y = 3 when x is less than 2, and equation y = 1 when x is greater than 2.

Answer:
**... there is a point where the gradient is 0**

The graph disproves three of these answers. It has no undefined limits (it is continuous). If you were to integrate between the crossings, you would get a negative area (some number between 0 and -2), because the curve goes under the x-axis.

However, the graph is an example of Rolle's theorem. The point where the gradient is 0 is exactly in the middle of A and B, at the co-ordinate (0, -1). Before this point, the gradient has been negative, and after it, the gradient will be positive. But at this point exactly, the gradient is not negative or positive: it is 0 exactly.

In question 4, we heard that points with a gradient of 0 are called "turning points", and here the graph is turning around from moving downwards to moving upwards. Rolle's theorem makes a bit more sense when we think of it like this: "if the function goes above or beneath the x-axis and later hits it again, then it has to turn around at some point". (There's a single exception if the graph remains on the x-axis exactly, where it makes a straight line that has gradient 0.)

Try to draw a function where the graph touches the x-axis twice but does not have any turning points. If you succeed in doing this, you've successfully found one of the conditions hinted at with "(specific type of)" in the statement of Rolle's theorem. Questions like "what exceptions could there be to this seemingly general rule?" make up the majority of what is studied and researched in pure Mathematics at university level. In this case, you will (spoiler!) find exceptions by thinking about discontinuities. For more sophisticated examples (where the function is continuous) you might like to research the terms "absolute value" and "Weierstrass function". Drop me a note if you have questions.

The graph disproves three of these answers. It has no undefined limits (it is continuous). If you were to integrate between the crossings, you would get a negative area (some number between 0 and -2), because the curve goes under the x-axis.

However, the graph is an example of Rolle's theorem. The point where the gradient is 0 is exactly in the middle of A and B, at the co-ordinate (0, -1). Before this point, the gradient has been negative, and after it, the gradient will be positive. But at this point exactly, the gradient is not negative or positive: it is 0 exactly.

In question 4, we heard that points with a gradient of 0 are called "turning points", and here the graph is turning around from moving downwards to moving upwards. Rolle's theorem makes a bit more sense when we think of it like this: "if the function goes above or beneath the x-axis and later hits it again, then it has to turn around at some point". (There's a single exception if the graph remains on the x-axis exactly, where it makes a straight line that has gradient 0.)

Try to draw a function where the graph touches the x-axis twice but does not have any turning points. If you succeed in doing this, you've successfully found one of the conditions hinted at with "(specific type of)" in the statement of Rolle's theorem. Questions like "what exceptions could there be to this seemingly general rule?" make up the majority of what is studied and researched in pure Mathematics at university level. In this case, you will (spoiler!) find exceptions by thinking about discontinuities. For more sophisticated examples (where the function is continuous) you might like to research the terms "absolute value" and "Weierstrass function". Drop me a note if you have questions.

This quiz was reviewed by FunTrivia editor WesleyCrusher before going online.

Any errors found in FunTrivia content are routinely corrected through our feedback system.

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