A quadratic graph has the shape known as a parabola. Here is an example:
The graph shows how two quantities that can vary are related to one another. That is, it shows how the choice of a quantity on the horizontal axis leads to a particular quantity on the vertical axis.
For example, the number $2$2 on the horizontal axis is related to $0$0 on the vertical axis, and $3$3 on the horizontal axis is related to $2$2 on the vertical axis.
We see that $1.5$1.5 on the horizontal axis is related to $-0.25$−0.25 on the vertical scale and this appears to be the least value possible for the quantity displayed on this axis.
In the graph above, what is the value on the vertical axis when the horizontal value is $1$1? What vertical value corresponds to the horizontal value $4$4? What vertical value corresponds to the horizontal value $-1$−1?
We can write the horizontal and vertical corresponding numbers as coordinate pairs $(x,y)$(x,y). So, we have $(1,0)$(1,0), $(4,6)$(4,6) and $(-1,6)$(−1,6).
Quadratic graphs arise when we have two quantities, we could call them $x$x and $y$y, that are related in a way that involves multiplying $x$x by itself to obtain the corresponding value of $y$y.
The diagram represents five rectangles that have the same perimeter but different areas.
Suppose we have $100$100m of fencing material that we wish to use to make a rectangular enclosure. The sides of the enclosure could look like any one of rectangles in the diagram or like any rectangle in between. We note that although the total of the side lengths is fixed at $100$100m, the area enclosed by the fences varies.
We could have sides $49$49m, $49$49m, $1$1m, and $1$1m, adding up to $100$100m and the area would be $49\times1=49$49×1=49 m^{2}. Or, we could have sides $30$30m, $30$30m, $20$20m, and $20$20m, again adding up to $100$100m but in this case enclosing an area of $20\times30=600$20×30=600 m^{2}.
We could find many other possible measurements for the sides and then calculate the area in each case. The goal would be to plot a graph that shows how the area of a rectangle is related to its width. To find the measurements we would probably go through a thought process something like the following:
At this point, it may be apparent that all this can be said more clearly and in less space with the use of some mathematical symbols. We could let $T=100$T=100 be the total length of fencing material, let $W$W be the width and $L$L be the length of the various rectangles, and let $A$A be the area enclosed. Then we can write:
$100$100 | $=$= | $2W+2L$2W+2L |
$50$50 | $=$= | $W+L$W+L |
$L$L | $=$= | $50-W$50−W |
Now that we have an expression for the length $L$L in terms of the width $W$W, we can find an expression for the area $A$A in terms of $W$W:
$A$A | $=$= | $W\times L$W×L |
$=$= | $W\left(50-W\right)$W(50−W) | |
$=$= | $-W^2+50W$−W2+50W |
This formula allows us to calculate $A$A for each value of $W$W chosen from the range $0$0m to $50$50m. The graph looks like the following:
The graph has the parabola shape because, in the formula for area, two terms involving $W$W have been multiplied together making it a quadratic formula.
We can see from the graph that the maximum area is obtained when the width is $25$25m. For this measurement, the rectangle is actually a square, since $2\times25+2\times25=100$2×25+2×25=100.
Note that the area of a rectangle is zero when the width $W$W is zero and also when $w=50$w=50.
In a room of $n$n people, if everyone shakes hands with everyone else, the total number of handshakes is given by $H=\frac{n\left(n-1\right)}{2}$H=n(n−1)2.
Here is the graph of $H$H:
What are the values of $n$n where the graph of $H$H intercepts the horizontal axis? Write your answers on the same line, separated by a comma.
The vertex of this parabola occurs when $n=0.5$n=0.5, as indicated below:
Select all true statements:
The value of $n=0.5$n=0.5 is not valid since you can't have half a person.
When there are $0.5$0.5 people there are a negative number of handshakes.
The minimum number of handshakes that can occur is $0.5$0.5.
Here is the graph of $H$H with only the values corresponding to whole numbers $n\ge2$n≥2:
These points are the only ones with useful interpretations.
Why is the first useful point, closest to the origin, at $\left(2,1\right)$(2,1)?
It takes at least $2$2 people to make $1$1 handshake.
$1$1 person can only perform $2$2 handshakes.
An object is released $700$700 metres above ground and falls freely. The distance the object is from the ground is modelled by the formula $d=700-16t^2$d=700−16t2, where $d$d is the distance in metres that the object falls and $t$t is the time elapsed in seconds. This equation is graphed below.
When $t=0$t=0, we get $d=700$d=700.
What does this mean?
The object is initially at ground level.
Initially, the object has not fallen any distance.
$t=0$t=0 is the time when the object lands on the ground.
When the object hits the ground, time stops.
Which of the following statements are true? Select all that apply.
The graph is always decreasing.
The graph is increasing for some values of $t$t, and decreasing for others.
The graph is neither increasing nor decreasing.
The graph is always increasing.
What does your answer to the previous part mean?
The object only loses height until it hits the ground.
The object falls faster and faster until it hits the ground.
The object gains height until it reaches $700$700 metres above ground.
A rectangle is to be constructed with $80$80 metres of wire. The rectangle will have an area of $A=40x-x^2$A=40x−x2, where $x$x is the length of one side of the rectangle. The graph of this function is given below.
What are the $x$x-intercepts of this graph?
Enter each $x$x-value on the same line, separated by commas.
What is the $x$x-value of the vertex?
What does the height of the parabola at $x=20$x=20 represent?
The smallest rectangular area that can be fenced off with $80$80m of wire.
The largest side of a rectangle that can fence off $80$80m^{2} in area.
The largest rectangular area that can be fenced off with $80$80m of wire.
The smallest side of a rectangle that can fence off $80$80m^{2} in area.