Graphs Foundation Revision Card Answers | Maths Made Easy

# Graphs Foundation Revision Card Answers

## FG1 – Gradient and ?=??+?

Question: Find the equation of the straight line that passes through the points (-3, -7) and (2, 8)

Answer: Remember we just need to find two numbers to describe our line: the gradient (m) and the y-axis intercept (c).

It’s easier if we start with the gradient. $m=\dfrac{\text{change in y}}{\text{change in x}}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{8—7}{2--3}=\dfrac{15}{5}=3$

So our equation so far is $y=3x+c$. To find c, we substitute any coordinates that the line passes through for x and y in this equation – let’s use the point (2,8).

$8=3\times 2 + c$

Now we just rearrange this equation to find c.

$8=6+c$

$c=2$

So our completed equation is $y=3x+2$

## FG2 – Coordinates and Midpoints

Question: Find the midpoint of the line segment that joins A and B as shown.

Answer: Point $A$ has coordinates $(-2, -2)$.

Point $B$ has coordinates $(0, 3)$.

By taking the average of the $x$ coordinates of $A$ and $B$, the $x$ coordinate of the midpoint is

$\frac{-2 + 0}{2} = -1$.

By taking the average of the $y$ coordinates of $A$ and $B$, the $y$ coordinate of the midpoint is

$\frac{-2 + 3}{2} = \frac{1}{2}$.

Therefore, the coordinates of the midpoint are $\left(-1, \frac{1}{2}\right)$.

## FG3 – Drawing Linear Graphs

Question: Plot the graph of 2?+1=8?. (hint: rearrange the equation so it’s in ?=??+? form!)

Answer: Let’s rearrange this equation. Subtract 1 from both sides:

$2y = 8x - 1$

Then, divide both sides by 2:

$y = 4x - \dfrac{1}{2}$

We can use this to find some coordinates. Since it’s a straight line we only really need two points – so we can pick any two sensible $x$ coordinates. Let’s use $x=0$ and $x=4$. We can then substitute these into the equation of the graph to give us the corresponding $y$ coordinates.

$y=4\times 0 -\dfrac{1}{2}=-\dfrac{1}{2}$

$y=4\times 4-\dfrac{1}{2}=15.5$

So we have the coordinates $(0,-\frac{1}{2})$ and $(4,15.5)$. Now we only need to plot these on a graph and join them with a straight line.

Another way to think about this is to think about what any $y=mx+c$ equation tells us. $m$ is the gradient, and $c$ is the y-axis intercept. So, for our equation, the $y$-intercept is $-\frac{1}{2}$, and the gradient is 4. This means that for every step we take in the $x$ direction, the $y$ coordinate increases by 4. This is all the information we need to plot the graph. The result should look like this:

## FG4 – Quadratic Graphs

Question: A curve is given by the function $y=x^2+4x-9$. Describe the shape of this curve. Plot the graph to check.

Answer: The coefficient of $x^2$ is positive, so the graph will be a U-shaped curve.

To plot the graph, we need to make a table of coordinates. First, we pick some x coordinates (e.g. -5 to 2), then we use the equation of the graph to work out the y-coordinates. For example, when x is -5, the y coordinate is given by $y=(-5)^2+4\times(-5)-9=25-20-9=-4$. We then do this with the other x coordinates until we have the following table:

Then we plot the coordinates $(-5,-4), (-4,-9)$ etc. to get the following graph.

## FG5 – Cubic and Reciprocal graphs

Question: Sketch the graph of the equation $y=ax^3+bx^2+c$ on the axis shown, where ?, ? and ? are integers.

What is the name given to this type of function?

Answer: The name of this function is cubic.

The sketch should be any general cubic sketch, like the one shown below.

## FG6 – Parallel Lines

Question: Lines $L_1$ and $L_2$ are parallel. Given that $L_1$ has equation $y=-2x+3$ and $L_2$ passes through the point (1, 5), find the equation of $L_2$.

Answer: Given that $L_1$ and $L_2$ are parallel, we know that they have same gradient. By comparing $L_1$ to $y=mx+c$, we can see that the gradient of $L_1$ is -2.

This means that the gradient of $L_2$ is -2 as well, so it has the equation $y=-2x+c$. Once again, we just need to find c and we’ll have the complete equation! To find c, we substitute the coordinates of any point that the line passes through into the equation of the line. We’re told that the line passes through $(1,5)$, so we substitute $x=1$ and $y=5$ into the equation and solve it to find c:

$5=-2\times 1+c$

$c=7$

So our complete equation for $L_2$ is $y=-2x+7$

## FG7 Distance-Time Graphs

Question:Valentina is going for a bike ride. Below is a distance-time graph that describes her full journey. Work out:

a) how long she was stationary for

b) the total distance travelled during her journey

c) Her average speed in kilometres per hour between 17:15 and 17:45

$\text{Gradient } = \dfrac{20}{0.5} = 40\text{km/h}$.