Flashcards · J560 · Topic 7

Graphs of equations and functions flashcards

25 cards on Topic 7, subtopics 7.01 to 7.04. Pick a subtopic or study them all.

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The four quadrants: Anticlockwise from top right: 1st (+, +), 2nd (−, +), 3rd (−, −), 4th (+, −). On the x-axis y = 0; on the y-axis x = 0.
Shapes of y = x², y = x³ and y = 1/x: y = x²: U-shape, minimum (0, 0), symmetrical about the y-axis. y = x³: rising S through (0, 0), in quadrants 1 and 3. y = 1/x: two branches in quadrants 1 and 3, touching neither axis.
How do you find the turning point of a quadratic that factorises?: Find the roots; the turning point’s x-value is their mean, then substitute for y. Example: y = (x + 6)(x − 4): roots −6 and 4, x = −1, y = −25, so (−1, −25).
Turning point of y = (x + p)² + qHigher: (−p, q). Example: x² + 6x + 7 = (x + 3)² − 2, minimum (−3, −2).
Key facts about y = kˣ for k > 0Higher: Passes through (0, 1) since k⁰ = 1. Always above the x-axis, with asymptote y = 0. k > 1: growth; 0 < k < 1: decay.
Graphs of sin x, cos x and tan x (degrees)Higher: sin x: starts at (0, 0), max 1 at 90°, period 360°. cos x: starts at (0, 1), min −1 at 180°, period 360°. tan x: through (0, 0), period 180°, asymptotes at 90° and 270°.
How many solutions does sin x = 0.4 have for 0° ≤ x ≤ 720°?Higher: For 0 < c < 1 the line y = c crosses the sine curve twice in each 360°, so 4 solutions.
Equation of a circle centre the originHigher: x² + y² = r², radius r = √(the number on the right). x² + y² = 49 has radius 7.

Reading m and c from 2y = 6x + 1: Rearrange to y = mx + c first: y = 3x + 1/2. Gradient 3, y-intercept (0, 1/2).
Equation of the line with gradient m through (x₁, y₁): y − y₁ = m(x − x₁), then tidy up. Or substitute the point into y = mx + c to find c.
Gradient and intercept in context: V = 40 + 15t litres after t minutes: Gradient 15: the rate, 15 litres added per minute. Intercept 40: the starting value, 40 litres at the start.
Parallel lines: Parallel lines have equal gradients. A line parallel to y = 4x − 3 is y = 4x + c; use a given point to find c.
Perpendicular gradientsHigher: m₁ × m₂ = −1: flip the fraction and change the sign. 3 becomes −1/3; −2/7 becomes 7/2.
How do you find the tangent to x² + y² = r² at (a, b)?Higher: 1. Radius gradient = b/a. 2. Tangent gradient = −a/b (negative reciprocal). 3. Use y − b = m(x − a).

Adding a to the whole equation: y = (original) + aHigher: Translation a units up (a units down if it is − a).
Replacing x by (x + a) in an equationHigher: Translation a units left: the opposite way to the sign. Replacing x by (x − a) moves it a units right.
y = −(original) and replacing x by −xHigher: y = −(original): reflection in the x-axis (every y changes sign). x replaced by −x: reflection in the y-axis (every x changes sign).
y = (x + 2)² − 1: turning point and y-interceptHigher: y = x² moved 2 left and 1 down: turning point (−2, −1). y-intercept (0, 3).
y = 2ˣ − 3: asymptote and y-interceptHigher: y = 2ˣ moved 3 down: asymptote y = −3, y-intercept (0, −2).

Distance-time graph features: Up-sloping line: constant speed away. Horizontal: stationary. Down-sloping: constant speed back. Steeper means faster. Gradient = speed.
Average speed for a trip out and back: Total distance travelled ÷ total time. The distance travelled is not 0 just because you return to the start. Example: 30 km out and 30 km back in 6 h: 60 ÷ 6 = 10 km/h.
Conversion graphs: currency compared with Celsius to Fahrenheit: Currency: a straight line through the origin; gradient = exchange rate. Temperature: F = 1.8C + 32, a straight line not through the origin, so not a proportion graph.
Velocity-time graph: what do the gradient and area give?Higher: Gradient = acceleration (m/s²); a negative gradient is deceleration. Area under the graph = distance travelled.
Chord gradient compared with tangent gradientHigher: Chord (joins two points on the curve): average rate of change. Tangent (touches at one point): instantaneous rate of change; it is an estimate.
Estimating area under a curve with trapezia: over or under?Higher: If the curve bends so the trapezia fall short of it, the estimate is too small; if the trapezia lie above the curve, it is too big.