## FG1 – Gradient and 𝑦=𝑚𝑥+𝑐

**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

**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 2𝑦+1=8𝑥. (hint: rearrange the equation so it’s in 𝑦=𝑚𝑥+𝑐 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 𝑎, 𝑏 and 𝑐 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

**Answer:**

a) We can see that the graph was flat for the duration of one big square. From the axis, we can see that two big squares total 15 minutes, therefore one big square is worth 7.5 minutes, so she was stationary for 7.5 minutes.

b) Valentina travelled away 25km away from home, stopped briefly, and then travelled 25km back home. Therefore, she travelled 50km in total.

c) We need to calculate the gradient of the graph between 17:15 and 17:45. This period lasted for 30 minutes, which is equivalent to 0.5 hours – this is the “change in x”. During this period, she increased her distance from home from 5km up to 25km, meaning she travelled 20km in total – this is the “change in y”. So, we get

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

**Note: **if you are asked to calculate the average speed over a longer period of time which contains several lines of different graphs, you still want to do the same calculation: divide the change in distance by the change in time. Even though you aren’t strictly calculating the gradient of one particular line, it’s still as if you’re looking at calculating an average gradient during that period.