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Linear and simple quadratic graphs - Mathematics Year 9

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Introduction

Where does a graph cross an axis, and what is the difference between an exact root and an estimate? Build a straight line and a parabola from their equations, then compare intersections using a second method.

Produce two graphs with labelled intercepts. Estimate an additional irrational root to one decimal place, retain its precision, and explain why a curved graph cannot be replaced by straight segments between a few points.

Build a straight line and label its intercepts

A graph of a function contains points whose coordinates satisfy its rule. The horizontal coordinate is x and the vertical coordinate is y. A table records selected points; the function specifies the relationship between them. Use an ordered pair in the correct order, not a pair of interchangeable labels.

For the source line $y = 3 x - 6$ , each increase of one in x increases y by three. The gradient is 3 and the y-intercept is −6. The y-axis contains points with x=0, so substituting zero gives y=−6 and intercept(0, −6). The x-axis contains points with y=0, so solve 0=3 x −6 to obtain x=2 and intercept(2, 0). An intercept is a point; its single nonzero coordinate is often named as shorthand.

Build the first graph.

x	y=3 x −6
-1	-9
0	-6
1	-3
2	0
3	3

Line: y=3x−6
y
 3 | .  .  .  O 
 2 | .  .  .  . 
 1 | .  .  .  . 
 0 | .  .  O  . 
-1 | .  .  .  . 
-2 | .  .  .  . 
-3 | .  O  .  . 
-4 | .  .  .  . 
-5 | .  .  .  . 
-6 | O  .  .  . 
   +------------
     0  1  2  3  x
O = listed point; . = blank.

Draw axes spanning at least x 0–3 and y −6–3 with one-unit intervals. Plot(0, −6), (1, −3), (2, 0), (3, 3), then use a ruler to draw one straight line through them. Extend it within the chosen domain. The text schematic marks samples only; the table also supplies(−1, −9) if your axes extend further. The labelled equation establishes linearity rather than the mere visual alignment of a few points.

Check the gradient using two points: (3−(−6))/(3−0)=9/3=3. This independent difference calculation agrees with the coefficient of x. Confusing a negative intercept with a negative gradient would predict the wrong direction: the line rises as x increases even though it crosses the vertical axis below zero.

If x represents a number of purchased items, its permitted values might be nonnegative integers. An abstract real function has a different domain. Do not attach a realistic context to every point without stating units and restrictions. The bare source graph has mathematical coordinates; it is not an observed journey or a measured dataset.

Retain the equation, table, axis scale and two labelled intercepts. A sketch should show the important shape and features; an accurate plotted graph additionally places points consistently on its stated scale. Neither screen pixels nor a table of exact values proves that a learner physically plotted the graph accurately.

Build a parabola and justify its turning point

For the source parabola $y = x^{2} - 4$ , square the entire signed x value before subtracting 4. Thus x=−3 gives(−3)²−4=9−4=5. Replacing(−3)² by −9 would be the first invalid step. Positive and negative inputs of the same magnitude produce equal outputs.

Build the second graph.

x	y=x²−4
-3	5
-2	0
-1	-3
0	-4
1	-3
2	0
3	5

Parabola: y=x²−4
y
 5 | O  .  .  .  .  .  O 
 4 | .  .  .  .  .  .  . 
 3 | .  .  .  .  .  .  . 
 2 | .  .  .  .  .  .  . 
 1 | .  .  .  .  .  .  . 
 0 | .  O  .  .  .  O  . 
-1 | .  .  .  .  .  .  . 
-2 | .  .  .  .  .  .  . 
-3 | .  .  O  .  O  .  . 
-4 | .  .  .  O  .  .  . 
   +---------------------
    -3 -2 -1  0  1  2  3  x
O = listed point; . = blank.

Draw axes covering x −3–3 and y −4–5 at one-unit intervals. Plot all seven table points and draw a smooth U-shaped curve through them. Do not join adjacent samples with a ruler as separate straight segments: the function is quadratic between the listed inputs too. The schematic gives exact labelled sample positions but does not pretend that its blank grid cells are a smooth rendered curve.

The graph is symmetric about x=0 because(−x)²=x². Its minimum turning point is(0, −4): a real square is at least zero, and zero occurs at x=0. This argument justifies the lowest value for all real inputs, not only for seven sampled points. The curve falls towards this point from the left and rises to its right.

The y-intercept is(0, −4). Roots occur where y=0, so x²−4=0 means x²=4. Both x=−2 and x=2 work. Substituting each into the original function gives 0. Naming only 2 confuses the principal square root of 4 with all solutions of the equation x²=4. The x-intercepts are(−2, 0) and(2, 0).

Compare the two completed graphs. The line has a constant gradient; the parabola's steepness changes. A vertical shift of the square graph by −4 changes its output values and roots but retains its symmetry about x=0. An unlabelled U shape does not establish the correct turning point or intercepts: state the rule, axes and units or coordinate meaning.

Use a table to check plotting arithmetic and algebra to verify exact intercepts. A chosen scale can make a crossing look different on screen without changing the function. Keep exact coordinate claims separate from visual estimates and any actual hand-drawn work.

Find intersections and estimate roots with stated precision

Read intersections as simultaneous conditions. A shared point lies on both graphs, so both rules must give the same y for the same x. For y=x²−4 and y=3 x −6, equate the outputs: x²−4=3 x −6. Rearrange to x²−3 x+2=0 and factor to(x −1)(x −2)=0. The roots x 1 and x 2 give y −3 and y 0, so the common points are(1, −3) and(2, 0). Check both original rules at each point; one line check alone is insufficient.

Estimate a non-integer root. For a new graph y=x²−2, the positive root lies between 1 and 2. Use a finer table: at 1.4, the output is −0.04; at 1.5, it is 0.25. A sign change with this continuous quadratic places the crossing between them, but that interval alone does not establish the correct rounding to one decimal place. The midpoint 1.45 is the rounding boundary.

Refine: 1.41²−2=−0.0119 and 1.42²−2=0.0164. On positive x, the square function increases, so the root is between 1.41 and 1.42. This whole interval lies below 1.45; the positive root is therefore 1.4 to one decimal place. The negative root is −1.4 to one decimal place by symmetry. Neither rounded value is an exact solution: 1.4² is 1.96, not 2.

When plotting this local estimate, use x 1.40–1.45 with sufficiently fine subdivisions and a suitable y scale rather than claiming a tenth from a coarse graph. Label the result as approximate and state the precision. An algebraic bound checks whether the graph estimate is plausible; the course does not fabricate an actual measured reading from a learner's paper.

Compare the original Corbettmaths table-and-plot example y=x²+2 x −1 with our different y=x²−4. Predict a negative-input square carefully. The original preview shows a separate table; its values are not the source values for our two graphs. Original: corbettmaths, Drawing quadratics - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Transfer and error analysis. A new line y=2 x+4 has y-intercept(0, 4) and x-intercept(−2, 0). A new parabola y=(x −1)²−9 has minimum(1, −9) and roots −2, 4. Substitute these roots and compare the symmetry halfway between them. An answer “root −4” for y=x²−4 incorrectly substitutes x=0 and reports a y-intercept as a root.

Retain two labelled graphs, their tables, the exact source intercepts, both intersections and the one-decimal-place root estimate with bounds. Explain one first error and a new-value transfer. Record actual later retrieval separately; correct answers here do not establish unobserved practical graphing accuracy.

Interactive Tasks

Quiz: Test Your Knowledge

For y=3 x −6, the y-intercept is? (0, −6) (!−6, 0) (!2, 0) (!0, 3)

The x-intercept of that line is? (2, 0) (!0, 2) (!−2, 0) (!0, −6)

For y=x²−4 at x=−3, y is? (5) (!−13) (!−5) (!13)

Both roots of x²−4=0 are? (−2 and 2) (!Only 2) (!−4 and 4) (!Only −4)

The minimum point of y=x²−4 is? (0, −4) (!−4, 0) (!2, 0) (!0, 4)

Why not join quadratic samples with straight segments? (Its rule describes a smooth curve between samples) (!Every graph must be a straight line) (!Only positive x is allowed) (!Tables cannot contain negatives)

Both source-graph intersections are? (1, −3 and 2, 0) (!0, −4 and 0, −6) (!1, 3 and 2, 6) (!Only 0, 0)

Positive root of x²−2 to one decimal place? (1.4) (!1.5) (!2.0) (!1.0)

Is 1.4 an exact root of x²−2? (No; its square is 1.96) (!Yes; rounded values are exact) (!Yes; all roots are tenths) (!No; negative roots never exist)

For y=2 x+4, the x-intercept is? (−2, 0) (!2, 0) (!0, 4) (!0, −2)

Memory Game

Root	Input giving output zero
Y-intercept	Graph point where x is zero
Gradient	Change in y divided by change in x for a line
Turning point	Point where the curve changes rising or falling direction
Intersection	Point satisfying both plotted rules
Precision	Stated detail or rounding of an estimate

Drag and Drop

3	Gradient of the source line
0, −6	Source line y-intercept
0, −4	Source parabola minimum
−2 and 2	Exact roots of the source parabola
1.4	Positive root of x²−2 to one decimal place

Crossword Puzzle

ROOT	Input whose output is zero
PARABOLA	Shape of a nondegenerate quadratic graph
INTERCEPT	Point where a graph crosses an axis
GRADIENT	Line slope calculated from coordinate differences
SYMMETRY	Reflection relationship about the central line
ESTIMATE	Approximate value with stated precision

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Find source intercepts: Set x 0 for the y-intercept and y 0 for roots.
Complete both tables: Check all negative-input squares.
Explain the minimum: Use x²≥0 to justify 0, −4.
Distinguish exact and approximate: Check why 1.4² does not equal 2.

Standard

Produce both graphs: Use labelled axes, appropriate scales, straight line versus smooth curve and all source intercepts.
Solve both intersections: Equate outputs, factor and check both original rules.
Estimate a root to one decimal place: Use finer plotting and the 1.41–1.42 bound to justify rounding.
Correct an intercept error: Explain why −4 alone is not a root of x²−4.

Advanced

Transfer new graph features: Find line 2 x+4 intercepts and parabola(x −1)²−9 roots and minimum.
Compare algebra with graph reading: State exact coordinates versus scale-dependent estimates.
Challenge coarse rounding: Explain why the interval 1.4–1.5 alone cannot select one-decimal rounding.
Retain actual plotted evidence: Keep the two graphs, bounds, first error and real later retrieval.

Learning Check

Two plotted graphs: Retain the source tables, axes and intercepts.
Turning and symmetry reasoning: Justify the source quadratic minimum for all real x.
Source intersections: Solve and check both rules at each pair.
Root estimate and precision: Bound the positive root of x²−2 and round.
Unfamiliar transfer: Find the stated new graph features independently.
First error and evidence: Repair a root/intercept confusion and retain actual work.

Worked checks and assessment guidance

For y 3 x −6, use x −1/0/1/2/3 with y −9/−6/−3/0/3. Line intercepts 0, −6 and 2, 0. For yx²−4, use x −3..3 and y 5/0/−3/−4/−3/0/5. Draw a smooth curve with intercepts −2, 0; 2, 0; 0, −4.
x²≥0, so x²−4≥−4, with equality at x 0. Equal-magnitude signed inputs give equal outputs; axis x 0 and minimum 0, −4. Finite sampling alone is not the general proof.
x²−4=3 x −6 gives(x −1)(x −2)=0. At x 1, both y −3; at x 2, both y 0. Shared points 1, −3 and 2, 0.
At 1.41 output −0.0119; at 1.42 output 0.0164. Increasing positive square function bounds root between them, below 1.45. Positive root≈1.4 to one decimal place; negative≈−1.4.1.4²1.96 does not equal 2.
Line y 2 x+4 has intercepts 0, 4 and −2, 0. Parabola y(x −1)²−9 has minimum 1, −9 and roots −2/4; both roots square the shifted input to 9.
A root sets y 0; setting x 0 instead finds a y-intercept. Negative inputs are squared before subtracting 4. Retain two graphs, both original checks, stated approximate precision and an actual later response; no measured plotting accuracy is fabricated.

Learning evidence

Retain two labelled source graphs, their tables and intercepts, both exact intersections, a one-decimal root estimate with independent bounds, first-error correction and new-value transfer.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, number, ratio, fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

2D Cartesian Coordinates.svg, H Padleckas, Public domain. Unchanged original; caption and learning activities are original.

Illustration for the area of a rectangle.svg, Stephan Kulla, CC0. Unchanged original; caption and learning activities are original.