A Algebra · graphs — revision question pack

4 specification points · notes, questions, answers and worked methods

A14 · Plot and interpret graphs (including reciprocal and exponential graphs) and graphs of non-standard functions in real contexts, to find approximate solutions e.g. simple kinematic problems

Explanation

A graph represents the ordered pairs satisfying a rule. Use intercepts, turning points and asymptotes to describe its behaviour.
Reciprocal graphs such as $y=k/x$ have two branches and approach the axes without meeting them.
Higher tier: exponential graphs change by a constant multiplier for equal changes in $x$ .
To solve equations graphically, plot both relations on the same axes and read every intersection accurately.
In a real context, the examiner expects the coordinate to be interpreted using the quantities and units on the axes, not reported as an unexplained pair of numbers.

For

y=k/x

with

k>0

, reciprocal branches lie in opposite quadrants and approach both axes.

Worked example

The journey time $t$ hours is modelled by $t=24/v$ , where $v$ is the average speed in km/h. Find and interpret the point when $v=16$ .

1.Substitute $v=16$ into the model: $t=24\div16$ .
2.Evaluate to obtain $t=1.5$ , so the coordinate is $(16,1.5)$ .
3.Interpret the axes: at $16$ km/h, the journey takes $1.5$ hours.

Answer: $(16,1.5)$ ; travelling at $16$ km/h gives a journey time of $1.5$ hours.

Common mistakes

Don't read the $y$ -coordinate when the question asks for the solution in $x$ .
Don't give an intersection such as $(16,1.5)$ without interpreting either coordinate or its unit.

Exam tip

For an 'estimate' question, show the plotted curves and read the intersection to a precision justified by the graph scale.

Tier 1 · Easy

1. The time $t$ hours for a fixed journey is modelled by $t=24/v$ , where $v$ is the average speed in km/h. Work out the point on this graph when $v=16$ and interpret it.[2 marks]

Tier 2 · Standard

1. Higher only: Plot $y=2^x$ and $y=9-2x$ for $1\le x\le3$ . Use the intersection to estimate the solution of $2^x=9-2x$ to one decimal place.[4 marks]

Tier 3 · Hard

1. For $0\le t\le5$ , two moving objects have distances from a marker modelled by $d=3t^2+2$ and $d=14t$ , with $d$ in metres and $t$ in seconds. Draw both graphs and estimate the later time when the objects are equally far from the marker.[5 marks]

A15 · Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

Explanation

Higher tier only. Gradient is change in the vertical coordinate divided by change in the horizontal coordinate, with units formed from the axis units.
For a curve, draw a tangent at the required point and use two well-separated points on that tangent, not two points on the curve.
Estimate an area under a non-linear graph by dividing it into strips and applying the trapezium rule.
On a distance-time graph gradient represents speed; on a velocity-time graph gradient represents acceleration and area represents displacement.
The examiner expects a numerical result, correct units and an interpretation of its sign where relevant.

Worked example

A distance-time graph is a straight line from $(12,150)$ to $(32,510)$ , with time in seconds and distance in metres. Calculate and interpret its gradient.

1.Use change in distance divided by change in time: $\dfrac{510-150}{32-12}$ .
2.Evaluate $\dfrac{360}{20}=18$ .
3.Attach the units metres per second and interpret the constant straight-line gradient.

Answer: $18$ m/s; the object travels at a constant speed of $18$ m/s.

Common mistakes

Don't use two points on the curve instead of two points on the tangent when estimating a gradient.
Don't add trapezium heights without multiplying by half the strip width.
Don't report a velocity-time area in m/s instead of metres.

Exam tip

For an estimate, leave the tangent or trapezia visible because the method marks depend on the construction.

Tier 1 · Easy

1. A distance-time graph is a straight line from $(12,150)$ to $(32,510)$ , where time is in seconds and distance is in metres. Calculate and interpret its gradient.[3 marks]

Tier 2 · Standard

1. A velocity-time graph joins the points $(0,0)$ , $(6,15)$ , $(14,15)$ and $(20,3)$ with straight lines. Work out the distance travelled in the first $20$ seconds.[4 marks]

Tier 3 · Hard

1. A curved velocity-time graph passes through the values $v=4,9,18,32$ m/s at $t=0,2,4,6$ seconds. Use three trapezia to estimate the distance travelled. A tangent at $t=4$ passes through $(2,11)$ and $(6,25)$ ; estimate the acceleration then.[6 marks]

A16 · Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

Explanation

Higher tier only. A circle centred at $(0,0)$ with radius $r$ has equation $x^2+y^2=r^2$ .
Check that a given point lies on the circle by substituting its coordinates. The radius from the origin to a point $(a,b)$ has gradient $b/a$ when $a\ne0$ .
The tangent at that point is perpendicular to the radius, so its gradient is the negative reciprocal, and a point-gradient equation gives the tangent.
The examiner expects an exact equation, normally simplified, and it should pass through the stated point.
Horizontal and vertical radii produce vertical and horizontal tangents respectively.

The tangent at a point on a circle is perpendicular to the radius.

Worked example

The point $P(5,12)$ lies on $x^2+y^2=169$ . Find the equation of the tangent at $P$ .

1.The radius $OP$ has gradient $12/5$ , so the tangent gradient is $-5/12$ .
2.Use point-gradient form: $y-12=-\dfrac{5}{12}(x-5)$ .
3.Multiply by $12$ and rearrange to obtain $5x+12y=169$ .

Answer: $5x+12y=169$ .

Common mistakes

Don't use the radius gradient $12/5$ as the tangent gradient.
Don't find the negative reciprocal correctly but writes a line that does not pass through the given point.

Exam tip

Substitute the contact point into your final tangent equation for a quick accuracy check.

Tier 1 · Easy

1. Write down the equation of the circle with centre $(0,0)$ and radius $7$ .[1 mark]

Tier 2 · Standard

1. The point $P(5,12)$ lies on the circle $x^2+y^2=169$ . Determine the tangent's equation there.[4 marks]

Tier 3 · Hard

1. The tangent to $x^2+y^2=50$ at $Q(1,7)$ meets the positive coordinate axes. Find the exact area of the triangle enclosed by the tangent and the axes.[5 marks]

A17 · Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

Explanation

Keep an equation balanced by applying the same operation to both sides until the unknown is isolated. Expand brackets first, then collect unknown terms on one side and constants on the other.
With fractions, multiply every term by a common multiple of the denominators before simplifying.
A graphical solution is the $x$ -coordinate where graphs representing the two sides intersect, so its accuracy is limited by the graph scale.
The examiner awards method marks for valid algebraic steps; unexplained sign changes are not valid operations.
Check the solution by substituting it into the original equation and confirming both sides are equal.

Worked example

Solve $\dfrac{3(x-2)}{4}-\dfrac{x+1}{3}=\dfrac56$ .

1.Multiply every term by $12$ : $9(x-2)-4(x+1)=10$ .
2.Expand and collect terms: $9x-18-4x-4=10$ , so $5x=32$ .
3.Divide by $5$ : $x=\dfrac{32}{5}$ .

Answer: $x=\dfrac{32}{5}$ .

Common mistakes

Don't multiply only the fractional terms, not every term, when clearing denominators.
Don't change the sign of a term when moving it without performing the same addition or subtraction on both sides.

Exam tip

For a multi-mark 'solve' question, keep each balancing step visible so an arithmetic slip does not lose the method marks.

Tier 1 · Easy

1. Solve $7x+5=33$ .[2 marks]

Tier 2 · Standard

1. Draw $y=4.5x-2$ and $y=9.2-x$ on the same axes. Use the intersection to solve $4.5x-2=9.2-x$ , giving an estimate to one decimal place.[4 marks]

Tier 3 · Hard

1. Solve $\frac{3(x-2)}{4}-\frac{x+1}{3}=\frac56$ .[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

A14 · Plot and interpret graphs (including reciprocal and exponential graphs) and graphs of non-standard functions in real contexts, to find approximate solutions e.g. simple kinematic problems

Tier 1 · Easy

1. Answer
- $(16,1.5)$
- At an average speed of $16$ km/h, the journey takes $1.5$ hours.
Method: Substitute $v=16$ : $t=24/16=1.5$ . The coordinates are speed then time, so the graph contains $(16,1.5)$ and this represents a $1.5$ -hour journey at $16$ km/h.

Tier 2 · Standard

1. Answer
- $x\approx2.2$
Method: Draw the increasing exponential curve $y=2^x$ and the decreasing straight line $y=9-2x$ on the same axes. Their intersection has $x$ -coordinate about $2.2$ (the numerical value is about $2.201$ ), so the graphical estimate to one decimal place is $2.2$ .

Tier 3 · Hard

1. Answer
- $t\approx4.5$ seconds
Method: Plot the quadratic $d=3t^2+2$ and the line $d=14t$ . Equal distances occur at intersections. The later intersection has $t\approx4.52$ , so an appropriate graph gives about $4.5$ seconds; the earlier intersection near $0.15$ seconds is not requested.

A15 · Calculate or estimate gradients of graphs and areas under graphs (incl. quadratic and other non-linear); interpret e.g. distance-time, velocity-time and financial graphs (not calculus) [Higher only]

Tier 1 · Easy

1. Answer
- $18$ m/s
- The object travels at a constant speed of $18$ m/s.
Method: The gradient is $(510-150)/(32-12)=360/20=18$ . Distance divided by time has units m/s, so this is the object's constant speed.

Tier 2 · Standard

1. Answer
- $219$ m
Method: The distance is the area under the graph. From $0$ to $6$ seconds the triangle has area $\frac12\times6\times15=45$ . From $6$ to $14$ seconds the rectangle has area $8\times15=120$ . From $14$ to $20$ seconds the trapezium has area $\frac12(15+3)\times6=54$ . The total is $45+120+54=219$ m.

Tier 3 · Hard

1. Answer
- Estimated distance $=90$ m
- Estimated acceleration $=3.5$ m/s $^2$
Method: With strip width $2$ , the trapezium estimate is $\frac{2}{2}[4+2(9)+2(18)+32]=90$ m. The tangent gradient is $(25-11)/(6-2)=14/4=3.5$ . On a velocity-time graph this gradient is acceleration, so the estimate is $3.5$ m/s $^2$ .

A16 · Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

Tier 1 · Easy

1. Answer
- $x^2+y^2=49$
Method: Use $x^2+y^2=r^2$ with $r=7$ . Since $7^2=49$ , the equation is $x^2+y^2=49$ .

Tier 2 · Standard

1. Answer
- $5x+12y=169$
Method: The radius $OP$ has gradient $12/5$ , so the tangent gradient is $-5/12$ . Using point-gradient form gives $y-12=-\frac{5}{12}(x-5)$ . Multiplying by $12$ and rearranging gives $5x+12y=169$ .

Tier 3 · Hard

1. Answer
- $\frac{1250}{7}$ square units
Method: The tangent at $(1,7)$ is $x+7y=50$ . Its intercepts are $(50,0)$ and $(0,50/7)$ . The enclosed right triangle therefore has area $\frac12\times50\times\frac{50}{7}=\frac{1250}{7}$ square units.

A17 · Solve linear equations in one unknown algebraically (including those with the unknown on both sides of the equation); find approximate solutions using a graph

Tier 1 · Easy

1. Answer
- $x=4$
Method: Subtract $5$ from both sides to get $7x=28$ . Divide both sides by $7$ , giving $x=4$ .

Tier 2 · Standard

1. Answer
- $x\approx2.0$
Method: The graphical solution is the $x$ -coordinate of the intersection. The lines meet at about $(2.04,7.16)$ , so the requested estimate is $x\approx2.0$ to one decimal place.

Tier 3 · Hard

1. Answer
- $x=\frac{32}{5}$
Method: Multiply every term by $12$ : $9(x-2)-4(x+1)=10$ . Expanding gives $9x-18-4x-4=10$ , so $5x=32$ and $x=32/5$ .