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This page sets out the method for drawing quadratic graphs in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Drawing quadratic graphs topic page.
When you see 'complete the table of values and draw the graph', work one column at a time.
When asked to solve an equation or find the turning point using the graph.
On the curve y=x^2-4, the y-value when x=3 is 5.
The roots of the curve y=x^2-4 are -2 or 2.
The y-intercept of y=x^2-4 is -4, the value when x is 0.
The curve y=(x-2)(x-3) crosses the x-axis at 2 or 3.
Work out x^2, then add the constant, including a negative constant.
Roots are x-values. The intercept asked here is a y-value.
Substitute each x-value on its own line. For y=x^2-2x-3 at x=-1, write (-1)^2-2×(-1)-3=1+2-3=0. The square of a negative is positive. Getting -4 for (-2)^2 breaks the symmetry of the whole table, which is how you notice it before you plot.
A useful set for this curve is x=-1, 0, 1, 2, 3. The y-values are 0, -3, -4, -3, 0. The row reads the same forwards and backwards around x=1. That mirror is the line of symmetry. If one cell breaks the mirror, recalculate that cell. Do not "smooth it out" by plotting a wrong point.
Method: 1. Work out each x-value on its own line of working. 2. For y=x^2-2x-3 at x=-1, calculate (-1)^2-2×(-1)-3=0. 3. Square a negative number to a positive number before you carry on.
Plot the points with small crosses and draw one smooth curve through them. Do not join them with a ruler, and do not flatten the bottom into a horizontal segment. A quadratic turns. The pencil stays on the page and the curve keeps bending the same way between the points. A sharp corner is a plotting error or a straight-edge join.
The roots of y=x^2-2x-3 are the crossings x=-1 and x=3. Give both if the curve crosses twice. The turning point is halfway between the roots, at x=1, and the table says y=-4 there. So the turning point is (1, -4) and the line of symmetry is x=1. Writing "1" with no equation loses the form the question asked for.
Method: 1. Mark each calculated point with a small cross, then join the crosses as one smooth curve. 2. Do not join the points with a ruler, and do not flatten the turning point into a horizontal segment. 3. Keep the pencil on the page so the curve bends the same way between the plotted points.
To solve x^2-2x-3=5 from the graph of y=x^2-2x-3, draw the horizontal line y=5 and read the x-coordinates where it meets the curve. Algebra confirms the reading. Subtract 5: x^2-2x-8=0, so (x-4)(x+2)=0. The solutions are x=4 and x=-2. Both should sit on your curve at height 5.
Show the reading on the graph with a dotted line from the intersection down to the x-axis. A correct algebraic answer with no mark on the graph can lose the "use your graph" mark. A reading that disagrees with the factorisation means a point in the table is wrong, usually a negative square. Recalculate that one cell, plot it again, and only then trust the crossing.
Method: 1. To solve x^2-2x-3=5 from the graph, draw the line y=5 and read the x-coordinates where it meets the curve. 2. Check that reading with algebra. 3. Subtract 5 to reach x^2-2x-8=0, which factorises as (x-4)(x+2)=0.
A sketch question gives you an equation but no grid, and asks you to show where the curve meets the axes and where its turning point is. You do not need a table of values. Find the shape first: a positive x^2 term gives a U and a negative one gives an upside-down U. Then find the three key facts, and label them with coordinates on your sketch.
For the y-intercept put x = 0. For the x-intercepts put y = 0 and solve by factorising. The turning point lies exactly halfway between the two roots, so its x-coordinate is the average of the roots. Substitute that value into the equation to find y. For y = x^2 - 2x - 8 the roots are -2 and 4, the turning point has x = 1, and y = 1 - 2 - 8 = -9.
Method: 1. Decide whether the curve is a U or an upside-down U from the sign of x^2. 2. Put x = 0 to find the y-intercept. 3. Put y = 0 and factorise to find where it crosses the x-axis. 4. Average the roots for the turning point x, then substitute to find y.
Tip: Write each point as a coordinate pair on the sketch. Marks are given for labelled points, not for neat curves.
A quadratic written as y = (x + p)^2 + q shows its turning point immediately. The bracket is zero when x = -p, and then y = q. So the turning point is (-p, q). Notice the sign flip on the first coordinate. For y = (x - 3)^2 + 2 the turning point is (3, 2), not (-3, 2). The point is a minimum for a U shape.
If you are given y = x^2 + 8x + 5, complete the square by halving the x coefficient: (x + 4)^2 gives x^2 + 8x + 16, which is 11 too many, so y = (x + 4)^2 - 11. The turning point is (-4, -11). For an upside-down U such as y = 7 + 6x - x^2, write it as 16 - (x - 3)^2, so the maximum is at (3, 16).
Method: 1. Halve the number in front of x to get p and write (x + p)^2. 2. Subtract the extra p^2 from the constant to find q. 3. Read the turning point as (-p, q). 4. Check by substituting x = -p into the original equation.
Table questions ask for two or three missing values in a row, usually including at least one negative x. Substitute each value in brackets: for y = x^2 - 4x at x = -1 write (-1)^2 - 4 × (-1) = 1 + 4 = 5. The sum of the two positive terms is where students get this wrong, writing 1 - 4 instead. A quick symmetry check helps: values equally far from the turning point are equal.
Real exams then ask you to plot the points and join them in a smooth curve, and later to read values from the graph. When you read, be exact with the scale. Look at what each small square is worth before you decide on a reading. In a context question, such as the height of a ball, the highest point of the curve is the maximum height, and the matching x value is the time.
Method: 1. Put each x value into brackets and work out the squared term first. 2. Write the whole line of working for each missing value. 3. Check the table is symmetrical about the turning point. 4. Check the scale before reading a value from the graph.
The question "By drawing a suitable straight line, solve x^2 - x - 6 = 0" gives you the curve y = x^2 - 2x - 3. Your job is to rewrite the equation so the left side is the curve. Here x^2 - x - 6 = 0 is the same as x^2 - 2x - 3 = 3 - x. The line to draw is y = 3 - x, and the crossing points give the solutions.
The method is always: write the curve on the left, then move everything else to the right. The right side is the line. If the equation has a constant on the right, such as x^2 - 2x - 3 = 5, the line is horizontal. If it has an x term, the line slopes. After drawing, read the x-coordinates of the crossing points only. You can confirm with algebra by solving the quadratic exactly.
Method: 1. Write the equation you must solve and the equation of the curve. 2. Rearrange so the curve expression is on one side. 3. Read the other side as the equation of the line. 4. Draw the line, read the crossing x values, and check them in the equation.
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