HookHow a man falling from space is one velocity–time graph
On 14 October 2012 Felix Baumgartner stepped off a capsule about \(39\) km above New Mexico and fell. For the first thirty seconds he accelerated almost as if in a vacuum, because at that altitude there is barely any air to push back, and he became the first human to break the sound barrier in freefall, hitting roughly \(1{,}358\) km/h — about Mach \(1.25\). Then, as he dropped into thicker air, the drag built, his acceleration fell to zero, and he settled to a steadier speed before his parachute opened. Every part of that story is a single graph: plot his speed up the side and time along the bottom, and the whole jump becomes a curve you can read.
That curve is the heart of A2.2. Its gradient — how steeply speed climbs — is his acceleration, near \(9.8\) m/s\(^2\) at the top and zero once drag balances gravity. The area underneath it is the total distance he fell. Where the curve levels off is a turning point in his acceleration. This section teaches you to turn a relationship into a picture and read the picture back: coordinates and straight lines, the parabola and where it turns, the family of standard curve shapes, how a graph shifts and flips when you change its equation, how to measure a rate and an area off a curve in a real context, and finally the circle and the tangent that just grazes it. Higher papers examine all of this on both calculator and non-calculator papers, and it quietly underpins kinematics, proportion and every rate-of-change question you meet later.
ModelCoordinates and the meaning of gradient and intercept
A coordinate \((x,\ y)\) is an address: go along the \(x\)-axis first, then up the \(y\)-axis — always in that order, because \((3,5)\) and \((5,3)\) are different points. The origin \((0,0)\) is where the axes cross, and they carve the plane into four quadrants, numbered anticlockwise from top-right. The whole of Higher graph work depends on placing a point in any quadrant without hesitating over the negatives: negative \(x\) means left of the origin, negative \(y\) means below it, so \((-4,1)\) sits top-left and \((4,-1)\) sits bottom-right.
Every straight line carries two numbers you must be able to read both from a picture and from its equation. The gradient \(m\) is the steepness, measured as \(\dfrac{\text{rise}}{\text{run}}\) — the change in \(y\) divided by the change in \(x\) between two points, using the axis scale and not the squares. A downhill line has a negative gradient. The y-intercept \(c\) is the value of \(y\) where the line crosses the \(y\)-axis, i.e. when \(x=0\). Written as \(y=mx+c\), the equation hands you both directly — but only once it is arranged into that form, so \(2y=6x+4\) has gradient \(3\), not \(6\), after dividing through by \(2\).
MechanismStraight lines — parallel, perpendicular, and finding the equation
Two versions of the same job appear constantly. Given two points, find the gradient with \(\dfrac{\text{rise}}{\text{run}}\), then substitute one point into \(y=mx+c\) to solve for \(c\). Given one point and a gradient, you already have \(m\), so go straight to the substitution.
The Higher content is the relationship between two lines. Parallel lines have equal gradients — they climb at the same rate and never meet, so a line parallel to \(y=3x+1\) has gradient \(3\) whatever its intercept. Perpendicular lines have gradients whose product is \(-1\): if one gradient is \(m\), the perpendicular gradient is the negative reciprocal \(-\dfrac{1}{m}\). So a line at right angles to \(y=3x+1\) has gradient \(-\dfrac{1}{3}\), because \(3\times\left(-\dfrac{1}{3}\right)=-1\). This is the single most examined new idea in the straight-line topic, and it links directly to the circle work later, where a tangent is perpendicular to a radius.
Find the equation of the line perpendicular to \(y=2x-5\) that passes through \((4,\ 1)\).
The given line has gradient \(2\). The perpendicular gradient is the negative reciprocal, \(-\dfrac{1}{2}\), since \(2\times\left(-\dfrac{1}{2}\right)=-1\). Now substitute the point \((4,1)\) into \(y=-\tfrac{1}{2}x+c\): \(1=-\tfrac{1}{2}(4)+c\), so \(1=-2+c\), giving \(c=3\). The equation is \(y=-\tfrac{1}{2}x+3\). Check the point lies on it: \(-\tfrac{1}{2}(4)+3=-2+3=1\) ✓. The whole answer turns on getting the negative reciprocal right — flipping the fraction but forgetting the minus is the classic slip.
ModelQuadratics — roots, intercept and the turning point by completing the square
A quadratic \(y=x^2+bx+c\) draws a symmetrical U-shaped parabola (it opens upwards when the \(x^2\) term is positive). Three features carry marks. The roots are where the curve crosses the \(x\)-axis, the values of \(x\) that make \(y=0\), found by factorising and setting each bracket to zero. The y-intercept is where it crosses the \(y\)-axis, found by setting \(x=0\). The turning point is the minimum of a U or the maximum of an \(\cap\).
At Higher the turning point comes from completing the square, which rewrites \(x^2+bx+c\) as \((x+p)^2+q\). The value of \(p\) is half the coefficient of \(x\), and the completed form reveals the turning point at \((-p,\ q)\) directly, because a squared bracket is smallest — zero — when its inside is zero. This same completed form solves the equation and, if \(q\) is negative, hands you the roots too, without any factorising.
For \(y=x^2-6x+5\), find the turning point and the roots by completing the square.
Halve the coefficient of \(x\): half of \(-6\) is \(-3\), so start with \((x-3)^2\). But \((x-3)^2=x^2-6x+9\), which is \(9\) too big, so subtract \(9\) and add back the \(+5\): \(y=(x-3)^2-9+5=(x-3)^2-4\). The bracket is smallest when \(x-3=0\), i.e. \(x=3\), giving \(y=-4\), so the turning point is \((3,-4)\).
For the roots, set \(y=0\): \((x-3)^2-4=0\), so \((x-3)^2=4\), hence \(x-3=\pm 2\). That gives \(x=5\) or \(x=1\). Notice completing the square delivered the exact turning point and the roots from one rearrangement — which is why examiners prefer it to guessing symmetry.
CaseThe shape zoo — knowing a graph by its silhouette
Higher expects you to recognise, sketch and match six families on sight. A linear graph \(y=mx+c\) is a straight line. A quadratic (with an \(x^2\) term) is a symmetrical parabola. A cubic (with an \(x^3\) term) has a stretched-S shape that rises, flattens and rises again, flipped for a negative cubic. A reciprocal graph \(y=\dfrac{1}{x}\) is two curves in opposite corners that hug the axes without ever touching them, because you can never divide \(1\) by a number to reach exactly zero, nor divide by zero at all — those axes are its asymptotes.
Two shapes are new at Higher. An exponential graph \(y=k^x\) (with \(k\gt 1\)) is the curve of compound growth: it starts just above the \(x\)-axis on the left, passes through \((0,1)\) because \(k^0=1\), then steepens forever — the mathematics of viral spread, compound interest and unchecked bacterial growth. The trigonometric graphs \(y=\sin x\), \(y=\cos x\) and \(y=\tan x\) are the wave shapes: \(\sin\) and \(\cos\) roll between \(-1\) and \(+1\) and repeat every \(360^\circ\), one a shifted copy of the other, while \(\tan\) climbs and breaks with vertical asymptotes every \(180^\circ\). Knowing these silhouettes turns a matching question into an instant tick.
MechanismTransformations — shifting and flipping a known graph
Once you know a curve \(y=f(x)\), you can produce a family of related curves by changing its equation, and each change has a fixed geometric meaning. The two translations: \(y=f(x)+a\) shifts the whole graph up by \(a\) (a vertical move, and intuitive), while \(y=f(x+a)\) shifts it left by \(a\) — a horizontal move that feels backwards, because the \(+a\) inside the bracket moves the graph in the negative direction. That counter-intuition is the examiner's favourite trap.
The two reflections: \(y=-f(x)\) flips the graph in the \(x\)-axis (every output changes sign, so up becomes down), while \(y=f(-x)\) flips it in the \(y\)-axis (every input changes sign, so left becomes right). A quick way to keep them straight: a change outside the function affects \(y\) and does the obvious thing; a change inside the bracket affects \(x\) and does the opposite of what it looks like.
The graph of \(y=f(x)\) has a minimum turning point at \((2,\ -3)\). Where is that turning point on \(y=f(x-1)+4\)?
Work the two changes separately. The \(f(x-1)\) part is a horizontal translation: \(x-1\) inside the bracket shifts the graph right by \(1\) (inside, so opposite to the sign), moving the \(x\)-coordinate from \(2\) to \(3\). The \(+4\) outside is a vertical translation up by \(4\), moving the \(y\)-coordinate from \(-3\) to \(1\). The new turning point is \((3,\ 1)\). Getting the horizontal direction the wrong way — sliding left to \((1,1)\) — is the mistake most candidates make, so always say aloud 'inside the bracket, opposite way'.
DataRates and areas — reading speed and distance off a graph
The payoff of the whole section is that a graph's gradient and the area beneath it are physical quantities you can name. On a distance–time graph the gradient is speed. On a velocity–time graph the gradient is acceleration, and the area underneath is the distance travelled — exactly the two readings that describe Baumgartner's fall. For a straight-line section you read the gradient with \(\dfrac{\text{rise}}{\text{run}}\) and find the area with triangles and rectangles.
When the graph is a curve, Higher asks you to estimate. To estimate the gradient at a point, draw a tangent — a straight line just touching the curve there — and measure its gradient; this gives the instantaneous rate, such as acceleration at one instant. To estimate the area under a curve, split the region into thin trapezia, work out each area and add them; more strips give a closer estimate. Because a tangent and a hand-drawn area are approximations, always state that your answer is an estimate — the honesty is itself part of the mark scheme.
A car's velocity–time graph rises in a straight line from \((0\text{ s},\ 0)\) to \((5\text{ s},\ 40\text{ m/s})\), then stays flat at \(40\) m/s until \(10\) s. Find the acceleration in the first phase and the total distance.
Acceleration is the gradient of the first section: \(\dfrac{40-0}{5-0}=8\) m/s\(^2\). Distance is the area under the whole graph. The first phase is a triangle of area \(\tfrac{1}{2}\times 5\times 40=100\) m; the second phase is a rectangle of area \(5\times 40=200\) m. Total distance \(=100+200=300\) m. If the acceleration phase had been a curve instead of a straight line — as in a real freefall — you would split those five seconds into trapezium strips and sum them to estimate the same area.
ModelCircles at the origin and the tangent that grazes them
A circle centred at the origin with radius \(r\) has the equation \(x^2+y^2=r^2\). This is just Pythagoras: any point \((x,y)\) on the circle is a distance \(r\) from the centre, and \(x\) and \(y\) are the two legs of a right-angled triangle with hypotenuse \(r\). So \(x^2+y^2=25\) is the circle of radius \(5\), and you can test whether a point lies on it by checking the sum.
The headline Higher skill is the tangent — the straight line that touches the circle at exactly one point. The key fact is geometric: a tangent is always perpendicular to the radius at the point of contact. So to find a tangent's equation you find the gradient of the radius from the origin to the point, take its negative reciprocal for the tangent's gradient, then use the point to find \(c\). This is where the perpendicular-gradient rule from the straight-line topic pays off in full.
Find the equation of the tangent to \(x^2+y^2=25\) at the point \((3,\ 4)\).
First confirm the point is on the circle: \(3^2+4^2=9+16=25\) ✓. The radius runs from \((0,0)\) to \((3,4)\), so its gradient is \(\dfrac{4-0}{3-0}=\dfrac{4}{3}\). The tangent is perpendicular to this radius, so its gradient is the negative reciprocal, \(-\dfrac{3}{4}\). Now substitute \((3,4)\) into \(y=-\tfrac{3}{4}x+c\): \(4=-\tfrac{3}{4}(3)+c=-\tfrac{9}{4}+c\), so \(c=4+\tfrac{9}{4}=\tfrac{25}{4}\). The tangent is \(y=-\tfrac{3}{4}x+\tfrac{25}{4}\). Every tangent question is this same three-step chain: radius gradient, negative reciprocal, then fit the point.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Graphs appear on both the non-calculator (Paper 1) and calculator (Papers 2 and 3) papers, and the working is marked method-first. For 'find the equation of a line' questions, write the gradient as a fraction before simplifying and show the substitution line when you solve for \(c\) — those are the method marks that survive an arithmetic slip. For anything perpendicular, write the negative-reciprocal step explicitly, because that is exactly where the marks and the mistakes cluster.
On quadratics, complete the square in full view: show the halved coefficient, the subtracted correction, and the final \((x+p)^2+q\), then read the turning point straight off — a stated turning point without the completed square rarely gets full method credit. For transformations, name the direction and say 'inside the bracket, opposite way' to avoid the horizontal-shift trap. On rate and area questions from a curve, always label your reading an estimate, draw the tangent or the trapezium strips on the graph so the examiner can see the method, and attach the units — m/s or m/s\(^2\) — since a missing unit is a routinely dropped mark. For circle tangents, run the fixed chain every time: radius gradient, negative reciprocal, then substitute the point, and confirm the point lies on the circle first.