The graph of a function is the set of all points whose co-ordinates (x, y) satisfy the function `y= f(x)`. This means that for each x-value there is a corresponding y-value which is obtained when we substitute into the expression for `f(x)`.
Since there is no limit to the possible number of points for the graph of the function, we will follow this procedure at first:
- Select a few values of x (at least 5)
- Obtain the corresponding values of the function and enter them into a table
- Plot these points by joining them with a smooth curve
However, you are encouraged to learn the general shapes of certain common curves (like straight line,parabola, trigonometric and exponential curves, which you'll come across in later chapters). It's much easier than plotting points and more useful for later!
Example 1
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A man who is `2\ "m"` tall throws a ball straight up and its height at time t (in s) is given byh = 2 + 9t − 4.9t2 m.
Graph the function.
Answer
We start at `t = 0` since negative values of time haveno practical meaning here.
We choose values `0.5` seconds apart (if we were to use `t = 1\ "s"` intervals, we would not see enough detail on the graph).
| t | 0 | 0.5 | 1 | 1.5 | 2 |
| h | 2 | 5.3 | 6.1 | 4.5 | 0.4 |
Graph of projectile (parabola).
This shape is called a parabola and is commonin applications of mathematics.
NOTE:
(1) This graph is height against time. The ball went straight up, not forward. (Our graph may give the impression the ball moved in the x- direction as well as up, but this was not the case.)
(2) We could have written the function in this example with h(t) rather than just h. The following 2 equations mean the same thing.
h = 2 +9t −4.9t2
h(t) = 2 +9t −4.9t2
Example 2
The velocity (in `"m/s"`) of the ball in Example 1 at timet (in s) is givenby
v = 9 − 9.8t
Draw the v-t graph. What is thevelocity when the ball hits the ground?
Answer
This is a straight line, since it is in the form
y = mx + c
See more on Straight Line.
Since we've recognized it is a straight line, we only need to plot2 points and join them. But we find 3 points, just to make sure we have the correct line.
| t | 0 | 1 | 2 |
| v | 9 | -0.8 | -10.6 |
Graph of `v` against `t` - straight line.
Our graph starts at `t = 0` (since negative time values have no meaning in this example).
For the first `0.918\ "s"`, the ball is going up (positive velocity - that is, the blue line is above the t-axis), but slowing down.
Thereafter, the ball is coming down towards the ground and getting faster (the portion where the blue line is below the t-axis).
The ball hits the ground at approx `t = 2.04\ "s"` (we cansee this from Example 1). The velocity when theball hits the ground from the graph we just drew is about `-11\"m/s"`. The graph stops at this point.
Our graph assumes the ball lands in sand and doesn't bounce.
Normally, as we have done here, we take velocity in the updirection to be positive.
Example 3
Graph the function y = x − x2.
Answer
(a) Determine the y-values for a typical set of x-values and write them in a table.
| x | −2 | −1 | 0 | 1 | 2 | 3 |
| y | −6 | −2 | 0 | 0 | −2 | −6 |
(b) Since `y = 0` for both `x = 0` and `x =1`, check what happens in between.
That is, for `x=1/2,`we find that`y=1/4`.
Graph of `y=x-x^2`, a parabola.
Note the curve continues beyond what is shown in the graph. This is just a general question and there are no practical limits for either the x- or y-values.
Example 4
Graph the function `y=1+1/x`
Answer
(a) Note: y is not defined for `x = 0`, due todivision by `0`
Hence, `x = 0` is not in the domain
(b) Draw up a table of values:
| x | `-4` | `-3` | `-2` | `-1` | `1` | `2` | `3` | `4` |
| y | `3/4` | `2/3` | `1/2` | `0` | `2` | `3/2` | `4/3` | `5/4` |
(c) We know something strange will happen near `x = 0` (since the graph is not defined there). So we check what happens at some typical points between `x = -1` and `x =1`:
when `x = −0.5,` `y = 1 + 1/(−0.5) = 1 − 2 = −1`
when ` x = 0.5,\ y = 1 + 1/(0.5) = 1 + 2 = 3`
(d) As the value of x gets closer to `0`, the points get closer to they-axis, although they do not touch it. The y-axisis called an asymptote of the curve.
(To convince yourself of this, plot points where `x = 0.4`, `x = 0.3`, `x = 0.2`, `x = 0.1` and even `x = 0.01`.)
Graph of `y=1+1/x`, a hyperbola. It's a discontinuous function.
There is another asymptote in this curve: `y = 1`, which is marked with a dashed line. Notice the curve does not pass through this value.
Example 5
Graph the function `y=sqrt(x+1)`
Answer
(a) Note: y is not defined for values of x lessthan `-1`. (Try some in your calculator, like `x = −4`.)
(b) We determine the some x-values and the corresponding y-values and write them in a table:
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 0 | 1 | 1.4 | 1.7 | 2 | 2.2 | 2.4 |
(c) We draw our graph as follows.
Graph of `y=sqrt(x+1)`. It's a half-parabola.
Example 6
The electric power P (in watts) delivered by abattery as a function of the resistance R (in ohms) is :
`P=(100R)/(0.50+R)^2`
Plot the power as a function of the resistance.
Answer
(a) Negative values for R have no physicalsignificance, hence P is not plotted for negative valuesof R.
(b) Draw up a table of values:
| R | 0 | 1 | 2 | 3 | 4 | 5 |
| P | 0 | 44.4 | 32.0 | 24.5 | 19.8 | 16.5 |
(c) Check what happens between `R = 0` and `R = 1`ie.
when `R = 0.25`, `P = 44.4`
`R = 0.50`, `P = 50`
`R = 0.75`, `P = 48`
Graph of `P=(100R)/(0.50+R)^2`.
Note the axes are labelled with R (resistance) and P (power).
(d) Conclusions:
(i) The maximum power of `50\ "W"` occurs when resistance `R = 0.5\ "W"`
(ii) P decreases as R increases beyond `0.5\ "W"`
Exercises
Graph the given functions
Q1. y = x3 − x2
Answer
(a) There are no restrictions on the values that x can take in this example, since it is a general question with no practical significance.
(b) Draw up a table of values:
| x | -1 | 0 | 1 | 2 | 3 |
| y | -2 | 0 | 0 | 4 | 18 |
Since `y = 0` when `x = 0` and `x = 1`, we examine what happens between those 2 x-values:
When `x = 1/2, y = -1/8.`
Here is our graph:
Graph of y = x3 − x2, a cubic.
Q2. `y=sqrt(x)`
Answer
We can only take the square root of a positive number so `x ≥ 0`. The square root of a number can only be positive, so `y ≥ 0`.
This graph is actually one half of a parabola, with horizontal axis.
Graph of `y=sqrt(x)`, a half parabola.
Conical water tank
Q3. (Application) Water flows out of a tank in the shape of an inverted cone (i.e. the water flowsthrough the pointy end of the cone and the widest part of thecone is at the top). The volume of the water is decreasing at aconstant rate.
Draw a sketch graph of the height of the water in the coneversus the time.
Answer
We need to model the height at time t based on what we know about cones. We also need to assume several things. (We make life easy for ourselves as we go along. We are allowed to do this since we just need to come up with a basic graph for the height of the water at time t).
Intuitively, we expect the water height to decrease slowly at first, then to drop more quickly near the end.
Thevolume of a cone is
`V=(pir^2h)/3 `
For simplicity, let r = h, then
`V=(pih^3)/3 `
So the height of the cone (solving for h) is given by
`h=((3V)/pi)^(1"/"3)`
We take a cone with "easy" values, say `h = r = 10`. This has volume `1000π/3\ "units"^3`.
If the water drains out in `10` seconds, it means `100π/3\ "units"^3`will drain out each second (This is just `1/10` of the volume). Thus the amount of water left after t seconds is given by
Volume
`= 1000(π/3) − 100t(π/3) text( units)^3`
`= (1000 − 100t)π/3 text( units)^3`
The height at time t will be
`h=((3(1000-100t))/pixxpi/3)^(1"/"3) `
That is,
` h=(1000-100t)^(1"/"3) `
The graph of our model is given below.
Graph of `h=(1000-100t)^(1"/"3)`.
