Recognise a line or find points for its graph

Which line matches the equation?

At zero, the solid line covers the sloping dashed comparison line. Move the constant slider through negative and positive values to separate them. Compare their directions and vertical-axis crossings. The dashed horizontal and vertical step shows the changes between the two marked points

As the constant increases, what happens to the solid line?

Start at the intercept, then use the gradient

Use the gradient and intercept you read in the previous topic. Two distinct points determine the line

1. 1First point Put the intercept on the vertical axis \(\displaystyle P=(0,c)\)
2. 2Second point Write the gradient as a signed vertical change over a positive horizontal change. Move right by the denominator, then up or down by the signed numerator \(\displaystyle \begin{aligned}m&=-\frac{3}{4}\\P&=(0,2)\\Q&=(4,-1)\end{aligned}\)
3. 3Recognise or sketch Check that the candidate passes through both points. On paper, plot both points on labelled axes and extend a ruled line through them
4. 4Rearrange first when needed Make the vertical variable the subject before using this method. A vertical line is handled by its fixed horizontal coordinate

Gradient

\(\displaystyle m=-\frac{2}{3}\)

First point

\(\displaystyle P=(0,5)\)

Second point

\(\displaystyle Q=(6,q)\)

Find the vertical coordinate of the second point

Compare the horizontal move with the gradient's denominator, then scale the signed vertical change by the same amount

Which line matches the equation?

Which is the first incorrect line?

Construct the whole graph

Draw complete lines or supported segments on labelled axes with suitable numerical scales and all required points. Compare each drawing with its model

1. 1Reuse the worked method The reference example rearranges the equation, finds two points, verifies them and shows the resulting line. Use the same method for a different equation below \(\displaystyle 3y+4x=15\quad (0,5),\ (3,1)\)
2. 2Check coordinates, not appearance alone A different valid window or unequal axis scales can change the apparent angle. Check coordinate values and rise divided by run, not a memorised slant

From equation to table and complete graph

For 2y+3x=16, make a table at x=−2,0,2,6 and draw the complete line for all real x. Choose and label suitable axes and uniform numerical scales. Label both intercepts exactly and one further point. Verify two distinct plotted points in the original equation and explain why the graph continues beyond the four table entries

Review the equation, table and drawing

Rearranging gives y=8−3x/2. The ordered pairs are (−2,11),(0,8),(2,5),(6,−1). The intercepts are (0,8) and (16/3,0); keep the fractional intercept exact. A suitable window is −3≤x≤7 and −3≤y≤13, but other clear scales are valid. Plot distinct points, use a ruler and extend the line both ways: the table samples do not restrict the stated real domain. The domain and range are both R.

1. 1Verify in the given equation At (0,8), 2×8+3×0=16. At (2,5), 2×5+3×2=16. Moving two units right lowers y by three, so the gradient is −3/2.
2. 2Check the whole drawing Your line must pass through every calculated point, not only an intercept. Label the fractional intercept at its correct position between five and six on the horizontal axis

Compare your drawing with this model. A student instead joins (0,8) to (2,11). What is wrong?

Construct the special cases

On separate labelled axes, draw y=−2, x=3 and y=3x/2. For each, mark two distinct exact points and state its gradient, domain and range. Explain why the two intercept calculations for y=3x/2 give only one point, and why x=3 cannot be written as y=mx+c with a finite m. Show each complete line rather than only two dots

Review all three special lines

For y=−2, join (0,−2) and (2,−2) horizontally and extend both ways: gradient 0, domain R, range {−2}. For x=3, join (3,0) and (3,2) vertically and extend both ways: undefined gradient, domain {3}, range R. For y=3x/2, use (0,0) and (2,3), then extend the rising line: gradient 3/2, domain and range R.

1. 1Distinct points matter Both intercepts of y=3x/2 are the origin, not two distinct points. The extra point (2,3) determines the direction
2. 2A vertical line is a relation At x=3, many y-values occur. Its horizontal change between distinct points is zero, so a finite rise/run gradient and a single-valued rule y=mx+c are impossible

Draw only the supported model

A tank initially contains 42 L and drains at a constant 3 L per minute until empty. Model volume V litres after t minutes. Write the rule and physical domain, make a table at t=0,4,10,14, and draw the complete supported graph with units and endpoints. Recover the rule from two points on your graph. Explain whether connecting points is appropriate when time varies continuously, and what changes if a table merely records observations once per whole minute

Review the continuous model and sampled table

The model is V=42−3t for 0≤t≤14, with range 0≤V≤42. The table is (0,42),(4,30),(10,12),(14,0). Join the included endpoints (0,42) and (14,0) with the straight segment and do not extend it to negative volume or negative time. A possible display window is −1≤t≤16, −4≤V≤48; a display window is not the physical domain

1. 1Recover the equation The slope between the endpoints is (0−42)/(14−0)=−3 L/min, and the initial volume is 42 L. This recovers V=42−3t
2. 2Model versus observations The constant-rate physical model includes fractional minutes, so its graph is a continuous segment. Observations made only at whole minutes are discrete recorded points; drawing a continuous model through them adds the stated constant-rate assumption. Sampling does not change the domain of that separately stated model

The display window includes 16 minutes. Does this make a volume of negative 6 L a valid prediction from the draining model?