 Graphs Higher Revision Card Answers

HG1 – 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 $y=3x+2$

HG2 – 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:

$\dfrac{-2 + 0}{2} = -1$

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

$\dfrac{-2 + 3}{2} =\dfrac{1}{2}$

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

HG3 – Parallel and Perpendicular Lines

Question: Lines $𝐿_1$ and $𝐿_2$ intersect at right angles at the point (-2, 3). $𝐿_1$ has the equation $𝑦=−3𝑥−3$. Find the equation of $𝐿_2$.

Answer: Given that $L_1$ and $L_2$ are perpendicular, their gradients must follow the relationship $m_1 \times m_2 = -1$. Rearranging this equation tells us that the gradient of $L_2$ is given by $m_2=\dfrac{-1}{m_1}=\dfrac{-1}{-3}=\dfrac{1}{3}$.

So our equation for $L_2$ takes the form $y=\dfrac{1}{3}x +c$. To find c, we substitute the coordinates of any point that the line passes through into this equation. We’re told that the lines intersect at $(-2,3)$, so we can substitute $y=3$ and $x=-2$ into this equation and solve for $c$:

$3=\dfrac{1}{3}\times -2+c$

$3=-\dfrac{2}{3}+c$

$c=3\dfrac{2}{3}=\dfrac{11}{3}$

So our equation for $L_2$ is $y=\dfrac{1}{3}x +\dfrac{11}{3}$

HG4 – Quadratic Graphs

Question: A curve is given by the function $𝑦=𝑥^2+4𝑥−9$. Describe the shape of this curve, and find the coordinates of the points where it crosses the x-axis. Plot the graph to check. Answer: The coefficient of $x^2$ is positive, so the graph will be a U-shaped curve.

To find the points where the graph intersects the x-axis, we need to solve the equation $x^2+4x-9=0$. This equation is difficult to factorise so we’ll use the quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-4\pm\sqrt{4^2-4(1)(-9)}}{2(1)} = \dfrac{-4\pm\sqrt{52}}{2}= 1.6, -5.6$

So the coordinates of the intersections with the x-axis are $(1.6,0)$ and $-5.6,0)$

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. HG5 – Cubic, Reciprocal and Exponential Graphs

Question: Complete the table below for the equation $𝑦=𝑥^3+3𝑥^2−4$. Use this table to plot a graph of this equation.  What is the name given to this type of function? Answer: The name given to this type of function is cubic.

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.

We will complete this table by substituting in the values of $x$ to our equation to get the missing values of $y$. For example, when $x=-2$,

$y=(-2)^3+3(-2)^2-4=-8+12-4=0$

Continuing this with the rest of the $x$ values, we get the completed table below. Plotting the coordinates gives us this graph: HG6 – Turning Points of Quadratic Graphs

Question: Sketch a graph of the function $𝑦=𝑥^2+5𝑥+6$. Label the coordinates of the turning point and any intersections with the x-axis. Answer: Firstly, we must find the roots of this quadratic by factorising it and setting it equal to zero. Observing that $2\times3=6$ and $2+3=5$, we get that

$x^2+5x+6=(x+2)(x+3)$

Therefore, to find the roots we set

$(x+2)(x+3)=0$

This clearly gives two roots: $x=-2$ and $x=-3$. Then, to find the coordinates of the turning, we need the halfway point between the roots, which is

$\dfrac{-2+(-3)}{2}=-2.5$

This is the $x$ coordinate of the turning point. To find the $y$ coordinate, we put this value back into the equation to get

$y=(-2.5)^2+5(-2.5)+6=-0.25$

Then, the resulting sketch of the graph should look like Note: because the question asked you for a sketch, it doesn’t have to be a perfect drawing, it just has to have the correct shape and correctly identified & labelled roots and turning points.

HG7 – Graph Transformations

Question: The graph of $𝑓(𝑥)=𝑥^2−𝑥−6$. Sketch, on the same axes, the graphs of $𝑦=𝑓(𝑥)$ and $𝑦=−𝑓(𝑥)+4$. Answer: Firstly, $x^2 - x - 6$ factorises to $(x - 3)(x + 2)$, so it is a positive quadratic with roots at 3 and -2.

Now, $-f(x) + 4$ is both a reflection in the x-axis and a translation of 4 in the positive y-direction. We should do the reflection first and the translation second – it often helps to sketch the intermediate step to help you, and you can always rub it out afterwards. Here, the dotted line will be the intermediate step (the reflection before the translation). So, we get HG8 – Area under a graph

Answer: First, we can draw a straight line that approximately follows the graph. Then we can divide this into simple shapes – in this case, we can split it up into 1 triangle and 3 trapeziums. Next we work out the areas of each shape using the relevant shape area formulas.

A: $A=\dfrac{1}{2}a\times b = \dfrac{1}{2}10\times 3=15$

B: $A=\dfrac{1}{2}(a+ b)\times h_v=\dfrac{1}{2}(3+4)\times 20 = 70$

C:$A=\dfrac{1}{2}(a+b)\times h_v= \dfrac{1}{2}(4+6.5)\times 17 = 85$

D: $A=\dfrac{1}{2}(a+ b)\times h_v= \dfrac{1}{2}(6.5+2)\times 13= 55.25$

Adding all of these together gives us an estimated total area of $15+70+85+55.25=225.25$.

HG9 Equations of Circles & Finding Tangents to Circles

Question: What is the equation of a circle with centre (0,0) and radius √5? Find the equation of the tangent to this circle at (−1,2).

Answer: Remember the form of the equation of a circle is $x^2+y^2=r^2$, where $r$ is the radius. The radius is $\sqrt(5)$, so $r^2=(\sqrt(5))^2=5$, so the equation of our circle is $x^2+y^2=5$.

To find the equation of the tangent, we first think about the radius that meets the tangent at (-1,2): We know that the line perpendicular to the tangent passes through (0,0) and (-1,2), so we can work out the gradient: $(2-0)/(-1-0)=2/(-1)=-2$. We know that the tangent is perpendicular to this, so the gradient of the tangent must be $\dfrac{1}{2}$ (because the tangents of perpendicular lines multiply to make -1, and $\dfrac{1}{2}\times-2=-1$.

So the equation of the tangent so far is $y=\dfrac{1}{2}x+c$.

To find $c$, we substitute the coordinates (-1,2) into the equation:

$2=\dfrac{1}{2}\times -1 +c$

Then solve to get $c=2.5$

So the complete equation of the tangent is $y=\dfrac{1}{2}x+2.5$

HG10 – Distance-Time Graphs

Question: Neil’s journey is shown in the distance time graph below.

At which time interval 12:00 – 12:30, 12:30 – 13:30, 13:30 – 16:30 or 16:30 – 18:30 was Neil’s speed greatest. First we need to work out how far Neil travels between each time period:

– 12:00 – 12:30, he travels from 0km away to 12km away;

remembering that speed = the gradient at the time.

$Gradient = \dfrac{12}{0.5}=24 kmph$

– 12:30 – 13:30, he travels from 12km away to 44km away;

$Gradient = \dfrac{44-12}{1}=32kmph$

– 13:30 – 16:30, he stays in one place;

$Gradient = \dfrac{0}{1}=0 kmph$

– 16:30 – 18:30, he travels from 44km away to 0km away.

$Gradient = \dfrac{-44}{2}=22 kmph$

From this we can see that the greatest speed is between 12:30 – 13:30.

HG11 – Velocity-Time Graphs

Question: The speed-time graph below describes a 50-second car journey. Work out the total distance travelled by the car and work out the maximum acceleration of the car during the 50 seconds. Answer: Before we get stuck in to answering this, let’s go through the journey described by the graph. Firstly, the car accelerated from 0 to 15m/s over the first 10 seconds (because the line is straight, the acceleration is constant). Then, the line is flat, meaning the car’s speed was not changing for 10 seconds – it was moving at constant speed. Next, the car accelerated up to 25m/s over the next 10 seconds, and finally it spent the last 20 seconds decelerating back down to 0m/s.

Now, we want to work out the distance travelled. On a speed-time graph, the distance travelled is the area under the graph. To work out the area under this graph, we will break it into 4 shapes: A, B, C, and D – two triangles, a rectangle, and a trapezium are all shapes that we can work out the area of. So, we get $\text{A}=\dfrac{1}{2}\times10 \times15= 75\text{m}$

$\text{B}=10\times15=150\text{m}$

$\text{C}=\dfrac{1}{2}(15+25)\times10=200 \text{m}$

$\text{D}=\dfrac{1}{2}\times20\times25=250\text{m}$

Therefore, the total distance travelled is: $75+150+200+250=675\text{m}$. Next up, the acceleration. To do this, consider that acceleration is a measure of how quickly something’s speed is increasing. Therefore, given that the gradient is a rate of change of $y$ (speed) with respect to $x$ (time), we can work out the acceleration by finding the gradient of the graph.

Now, the question asks for “maximum acceleration”, so we can rule out certain parts of the journey. Specifically, the part where the graph is flat – there is no acceleration here – and the part where the graph slopes downward – it is decelerating here, so can’t be the maximum acceleration.

It might be obvious to you which of the two sections of the graph is steeper, but it isn’t always, so we’ll work out both just to be sure. Firstly, the first 10 seconds: the car’s speed goes from 0 to 15m/s and it takes 10 seconds, so we get

$\text{acceleration between 0s and 10s = gradient}=\dfrac{15-0}{10-0}=1.5\text{m/s}^2$

Then, the other portion we’re interested in is between 20 and 30 seconds. During this period, the speed increases from 15 to 25, so we get

$\text{acceleration between 20s and 30s = gradient}=\dfrac{25-15}{30-20}=1\text{m/s}^2$

The first one is larger, so the maximum acceleration is $1.5\text{m/s}^2$.

HG12 – Inequalities on Graphs

Answer: We’re going to pretend that the inequalities are equations and plot them as straight lines. The first one will be the solid plot of the line $y=1$, the second will be a solid plot of the line $y=3$, the third will be a dashed plot of the line $x=0$, and the fourth will be a dashed plot of the line $y=x$.

Now, we want to shade the area that is above the line $y=1$, below the line $y=3$, to the right of the line $x=0$, and above the line $y=x$.

The resulting graph looks like: Labelling the lines is not necessary but you may find it be a helpful intermediate step. Or you might not, which is totally fine – be your own person.