Graphs of Equations

The Coordinate Plane #

Across first, then up — every point has an address.

Coordinates are an address: how far across from the origin, then how far up

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Two number lines crossed at zero make a plane. The crossing point is the origin.

The pair (4, 2) is an address: from the origin, 4 across, then 2 up.

Order matters: (2, 4) is a different point. Across comes first, up second.

The plane keeps going left and down. Left of the origin, across turns negative.

Below the origin, up turns negative: (2, −3) is 2 across and 3 down.

Simultaneous Equations #

The one pair that satisfies both.

Two equations meet at the one pair of values that satisfies both

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The first equation is y = x. Every point on this line makes it true.

The second equation is y = 4 − x, and its line is where that one holds.

The lines cross once, at (2, 2). That pair sits on both of them.

Put x = 2 and y = 2 into each: 2 = 2, and 2 = 4 − 2. Both equations hold.

y = mx + c #

m is the slope, c is where it crosses.

In y = mx + c the m is the gradient and the c is the y-intercept

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This line crosses the y-axis at 2, so the y-intercept is c = 2.

A bigger m is a steeper gradient. Here it climbs 2 for every 1 across.

Writing a Line as Ax + By = C #

Both intercepts fall straight out.

In Ax + By = C each intercept falls out by setting the other letter to zero

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Move the x term across and y = 6 − 2x becomes 2x + y = 6.

It is the same line in a new form. It meets the axes at (0, 6) and (3, 0).

Set x = 0 to find one crossing, and y = 0 to find the other.

To go back, undo the moves: get y alone, then divide both sides by its coefficient.

Point-Slope Form of a Line #

A point and a gradient write it for you.

The form $y-y_1=m\left(x-x_1\right)$y − y₁ = m(x − x₁) writes a line straight from one point and its gradient

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This line has gradient 2 and passes through (2, 3). Those two facts fix it completely.

Rise over run from (2, 3) to any other point on the line is 2. That is the form.

Expand and tidy: y = 2x − 1. It is the same line, written in two ways.

This is why the form is useful: the point and the gradient drop straight in.

Quadratic and Cubic Graphs #

Squaring and cubing turn lines into curves.

Squaring or cubing the input bends a straight line into a curve

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Squaring makes a U, because a negative squared comes back up.

Cubing keeps the sign, so it dives below the axis on the left.

The Average Rate of Change Over an Interval #

The gradient of the chord across the interval.

The average rate of change across an interval is the gradient of the chord over it

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A curve has no single steepness, so pick two points and join them by a chord.

Rise over run, exactly as for a line: a rise of 8 over a run of 4 gives 2.

The curve is gentler than the chord at first and steeper at the end. 2 is the average.

Change the interval and the answer changes: from 0 to 4 the chord climbs 1 per step.

Estimating a Gradient with a Drawn Tangent #

Touch the curve once, then read the line.

Lay a ruler to just touch the curve at the point, then read the drawn line like any line

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At P the curve has one steepness. A ruler laid to just touch there draws the tangent.

Read the tangent like any line: from (1, 0) to (3, 2), rise 2 over run 2 — gradient 1.

A drawn tangent is an estimate — a careful drawing comes close to the true gradient.

Reciprocal Graphs #

Never touching either axis.

A reciprocal graph never touches either axis because you cannot divide by zero

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As x grows the value shrinks, but it never quite reaches zero.

A bigger numerator pushes the whole curve further out.

The Graphs of y = axⁿ #

Even powers mirror, odd powers half-turn.

One family $y=a{x}^n$y = axⁿ: the power sets the shape, and a sum of power terms mixes the shapes

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For n = −2, $y=\frac{1}{x^2}$y = 1/x² rises steeply near zero; squaring keeps both arms above the axis.

A negative a flips the shape: $y=-x^2$y = −x² is the same curve reflected in the x-axis.

In a sum each term takes over in turn: $y=x^3-2x$y = x³ − 2x follows −2x near zero, $x^3$x³ far out.

The Graph of y = |x| #

Two straight rays meeting in a V.

The graph of y = |x| is a V, because the modulus turns every negative output back up

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y = |x| is a V: two straight rays meeting at the origin.

Right of zero the graph is y = x; left of zero the minus sends it back up.

y = |x − 2| is the same V with its corner at 2, where the inside reaches zero.

y = |2x − 4| turns at 2 as well, and climbs twice as fast on each side.

The corner sits where the inside reaches zero, so solve that to find it.

Taking the Modulus of a Whole Expression #

Every part below the axis folds up.

Taking the modulus of a whole expression reflects every part below the x-axis up above it

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$y=x^2-4$y = x² − 4 runs below the axis between −2 and 2, dipping to −4 in the middle.

$y=|x^2-4|$y = |x² − 4| sends that dip back up: the part below the axis is reflected.

At x = 0 the curve was −4, so the modulus graph is 4 — same size, other side.

The graph touches the axis at −2 and 2 without crossing, so a corner sits at each.

Keep every value above the axis, and send every value below back up to its size.

Replacing x with a Modulus #

The right-hand half is copied across.

Replacing x by |x| throws away the left half of a graph and mirrors the right half into it

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Start with $y=x^2-4x$y = x² − 4x, drawn for x from 0 up: a dip to −4 and back to the axis.

Replace x by |x| and −3 gives what 3 gives, so the two sides match in height.

$y=x^2-4|x|$y = x² − 4|x| is that right-hand half, copied across the y-axis as a mirror.

y = |x| − 2 and y = |x − 2| are different V shapes: the bars sit in different places.

The Graph of y = 1/f(x) #

Sketching a reciprocal graph from the graph of f.

Every height of $y=\frac{1}{f\left(x\right)}$y = 1/f(x) is the reciprocal of f(x), so zeros of f become asymptotes and large becomes small

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Where f(x) = 0, $\frac{1}{f\left(x\right)}$1/f(x) has no value: the zeros of $f\left(x\right)=x^2-4$f(x) = x² − 4, at x = −2 and x = 2, become vertical asymptotes.

Where f(x) is 1 or −1, so is $\frac{1}{f\left(x\right)}$1/f(x), so the two graphs meet at those heights. Its sign always matches f(x).

As x runs away from 0, f(x) grows without limit, so $\frac{1}{f\left(x\right)}$1/f(x) shrinks toward 0: the x-axis is an asymptote.

f has a minimum at (0, −4), so $\frac{1}{f}$1/f has a maximum at $\left(0,-\frac{1}{4}\right)$(0, −1/4): the turning point keeps its x, and its height becomes 1 ÷ (−4).

Intersections of Graphs #

The point that solves both at once.

Where two graphs cross is the pair of values that satisfies both

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Each line has its own equation. They meet once, and that point solves both.

Solving Inequalities from Graphs #

Read the answer off where one graph is below the other.

f(x) < g(x) wherever the graph of f is below the graph of g, so the crossings mark the ends of the answer

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The graphs of $y=x^2-4$y = x² − 4 and y = x + 2 cross at x = −2 and at x = 3.

$x^2-4<x+2$x² − 4 < x + 2 wherever the parabola is below the line: between the crossings, −2 < x < 3.

$x^2-4>x+2$x² − 4 > x + 2 where the parabola is above the line, on both sides: x < −2 or x > 3.

The crossings come from solving the equation, and $\le$≤ or $\ge$≥ keeps them in the answer.

The Equation of a Circle #

Pythagoras from the center, written down.

Every point on a circle is r from the center, and Pythagoras turns that into an equation

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Every point (x, y) on a circle sits the same distance r from the center (a, b).

Step across then up and a right triangle appears: legs x − a and y − b.

Pythagoras on that triangle is the equation of every point on the circle.

Center it at the origin and a and b disappear, leaving $x^2+y^2=r^2$x² + y² = r².

Read it backwards: ${\left(x+1\right)}^2+{\left(y-2\right)}^2=9$(x + 1)² + (y − 2)² = 9 has center (−1, 2) and radius 3.

Center and Radius by Completing the Square #

Fold the expanded form back into brackets.

Completing the square in x and in y turns an expanded circle back into center-radius form

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Multiplied out, the equation of a circle hides its center and its radius.

Gather the x terms, gather the y terms, and move the constant to the other side.

Complete the square twice, once in x and once in y. Each leaves an extra constant.

Move both extra constants to the other side and the center-radius form is back.

So the center is (2, −3) and the radius is 4, the square root of 16.

The Tangent to a Circle at a Point #

At right angles to the radius that reaches it.

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

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On $x^2+y^2=25$x² + y² = 25, draw the radius out to (3, 4). Its gradient is $\frac{4}{3}$4/3.

The tangent there meets the radius at a right angle — this is true at every point.

Perpendicular gradients multiply to −1: flip $\frac{4}{3}$4/3 and change its sign to get $-\frac{3}{4}$−3/4.

Point-slope through (3, 4) tidies to the tangent line 3x + 4y = 25.

Where a Line Meets a Circle #

Substitute, solve, and send each x back.

Substituting a line into a circle leaves a quadratic whose roots are the meeting points

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The line y = x + 1 cuts $x^2+y^2=25$x² + y² = 25. The question is where, exactly.

Put x + 1 in place of y and the circle becomes one equation in x alone.

Expand and gather: a plain quadratic, $x^2+x-12=0$x² + x − 12 = 0.

Factor to find each x, then substitute it into y = x + 1 to find its y.

The two answers are the two crossings: (−4, −3) and (3, 4).

The Equation of an Ellipse #

A circle stretched by a different amount each way.

An ellipse is a circle stretched by a different amount along each axis

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Divide the circle equation by 9: each term now has the same 9 in its denominator.

Give the two terms different denominators and the circle stretches into an ellipse.

Set y to 0 and x reaches $\pm 5$±5; set x to 0 and y reaches $\pm 3$±3 — the two square roots.

Swap the two denominators and the same ellipse turns upright: $\frac{x^2}{9}+\frac{y^2}{25}=1$x²/9 + y²/25 = 1.

Subtract the center inside each bracket, as the circle does: $\frac{{\left(x-2\right)}^2}{16}+\frac{{\left(y+1\right)}^2}{4}=1$(x − 2)²/16 + (y + 1)²/4 = 1.

The Equation of a Hyperbola #

One minus sign, two branches, two asymptotes.

Changing the plus in an ellipse to a minus opens the curve into two branches

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One sign changes. A plus closes the curve; a minus opens it into two pieces.

$\frac{x^2}{9}-\frac{y^2}{4}=1$x²/9 − y²/4 = 1 touches the x-axis at $\pm 3$±3 and leaves the middle empty.

Put x = 0 and you need $y^2=-4$y² = −4 — no real y squares to a negative, so no point sits there.

Far from the center each branch hugs a straight line it never touches: an asymptote.

Replace the 1 by 0 and the equation splits into the two lines, $y=\pm \frac{2x}{3}$y = ±2x/3.

Distance-Time Graphs #

Steepness on the graph is the speed.

On a distance time graph the steepness is the speed

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A straight climb means a steady speed: the graph rises 20 in every hour.

A flat stretch means the distance is not changing, so it has stopped.

Speed-Time Graphs #

Steepness is acceleration; the area is distance.

On a speed time graph the height is the speed, steepness is acceleration, and the area underneath is distance

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The height is the speed now. Flat at 60 means a steady 60 — moving, not stopped.

A climbing line means the speed itself is rising by 20 each hour — that is acceleration.

Back to the steady 60: in 2 hours it covers 60 × 2 = 120 — the shaded area.

Under a climbing line the area is a triangle: half of 4 × 80 = 160, the distance covered.

A Line in Graph, Table and Equation #

Any one form rebuilds the other two.

One straight line lives as an equation, a table and a graph, and any one rebuilds the rest

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y = 2x + 1 as a table: each y is double x plus 1. The gap of +2 per row is the gradient.

Plot the rows and they line up — the table was the graph, sampled at whole numbers.

Read the graph back: it crosses y at 1 and climbs 2 per step — y = 2x + 1 again.