1MA1 · A11–A14AlgebraFoundation & Higher

Quadratic Graphs

A quadratic function produces a parabola. Its roots, intercept and turning point connect the equation to a picture, allowing you to interpret, sketch and solve quadratic problems graphically.

55 min Pearson Edexcel GCSE Maths (1MA1)

The big picture

A quadratic graph is the visual version of a quadratic expression. Its roots solve $f (x) = 0$ , its $y$ -intercept is $f (0)$ , and its turning point gives a minimum or maximum. Reading the curve and manipulating the equation are complementary skills.

Edexcel graph questions test plotting accuracy, recognition, interpretation and approximation. Labels, a sensible scale and a smooth curve are part of the mathematics because they determine how reliably values can be read.

What you'll learn

Complete a table of values and plot a quadratic graph accurately
Identify roots, intercepts, the axis of symmetry and turning point
Sketch a parabola from its equation or factorised form
Use a graph to solve equations approximately
Interpret a quadratic graph in context

Plotting a parabola

Calculate ordered pairs from the equation, plot them accurately and join them with one smooth curve. A quadratic graph is not made from straight line segments.

Substitute negative values with brackets. For $y = x^{2} - 2 x - 3$ at $x = - 2$ , enter $(- 2)^{2} - 2 (- 2) - 3 = 5$ ; without brackets, a calculator may square only the 2 and mishandle the sign.

1Substitute using brackets:

y = (- 1)^{2} - 2 (- 1) - 3

2Calculate each term:

1 + 2 - 3

Answer

y = 0

Tip — Plot small crosses, label both axes and use a continuous smooth curve through the points.

Roots, intercept and turning point

The roots are where $y = 0$ , so they are the $x$ -coordinates of the $x$ -intercepts. The $y$ -intercept occurs when $x = 0$ and is usually read directly from the constant term.

The turning point is the minimum of an upward-opening parabola or maximum of a downward-opening one. The vertical line through it is the axis of symmetry.

1The factors are zero at

x = 2

and

x = - 4

, so these are the roots.

2The axis of symmetry lies midway between the roots:

x = (- 4 + 2) /2 = - 1

3Substitute

x = - 1

y = (- 3) (3) = - 9

4At

x = 0

y = (- 2) (4) = - 8

AnswerRoots

- 4

and

2

; turning point

(- 1, - 9)

;

y

-intercept

(0, - 8)

Why this works

Symmetry explains why the turning point lies halfway between two roots: matching heights occur at equal horizontal distances from the axis.

Sketching from algebra

The sign of the $x^{2}$ coefficient gives the direction: positive opens upward and negative opens downward. Factorised form exposes the roots; completed-square form exposes the turning point.

A sketch communicates structure rather than exact plotting. Mark known intercepts and the turning point, show the correct direction, and draw a smooth symmetric curve.

y = a (x - r_{1}) (x - r_{2})

Factorised form displays roots

r_{1}

and

r_{2}

y = a (x - p)^{2} + q

Completed-square form has turning point

(p, q)

1The completed-square form gives turning point

(3, 5)

2The coefficient

- 2

is negative, so the curve opens downward.

3Its axis of symmetry is

x = 3

AnswerA downward-opening parabola with maximum

(3, 5)

and axis

x = 3

Solving equations with graphs

To solve $f (x) = 0$ , read the $x$ -coordinates where the graph of $y = f (x)$ meets the $x$ -axis. To solve $f (x) = k$ , draw the horizontal line $y = k$ and read the intersection values.

To solve $f (x) = g (x)$ , draw both graphs and read the $x$ -coordinates of their intersections. Answers from a graph are estimates and their accuracy is limited by the scale and line thickness.

1Rewrite the target equation as

x^{2} - 5 x + 2 = 4

2On the existing graph, draw the horizontal line

y = 4

3Read the

x

-coordinates of the two intersections.

AnswerDraw

y = 4

and read both intersection

x

-values

Why this works

Rearranging the target equation to match an already-drawn curve avoids plotting a new quadratic; the remaining side becomes a straight reference line.

Think like an examiner

For plotting questions, show substitutions in the table; a correct method can earn credit after one arithmetic error.
Give every root or intersection visible in the requested range and quote only sensible graph-based accuracy.
A sketch should label roots and turning point where known, even when no scale is required.
When solving with an existing graph, state and draw the additional line before reading intersections.

Remember these

f (x) = 0

Roots occur at the

x

-axis.

x = \frac{r _{1} + r _{2}}{2}

Axis of symmetry midway between roots.

y = a (x - p)^{2} + q

Turning point

(p, q)

Watch out for these

Joining plotted points with separate ruler lines.

A quadratic is a smooth curve; draw one continuous parabola.

Reporting the root as a coordinate such as

(3, 0)

when asked to solve an equation.

The solution is the

x

-value, so report

x = 3

Assuming the constant term is an

x

-intercept.

It is the

y

-intercept because setting

x = 0

leaves the constant.

Stretch yourself

A parabola has roots $x = - 2$ and $x = 6$ and passes through $(0, - 24)$ . Find its equation and turning point.

Hint — Begin with $y = a (x + 2) (x - 6)$ and use the known point to find $a$ . The axis lies midway between the roots.

Questions students ask

Why does a quadratic graph sometimes have no roots?

Its parabola may lie entirely above or below the

x

-axis. In that case no real input makes the function equal zero.

How accurate should a graphical solution be?

Use the scale: normally quote to the nearest small grid division, often one decimal place. Extra digits imply precision the graph cannot provide.

Key takeaways

Plot quadratic points accurately and connect them with a smooth curve.
Roots occur at $y = 0$ ; the $y$ -intercept occurs at $x = 0$ .
Factorised and completed-square forms reveal different graph features.
Graphical equation solutions come from intersections and are approximate.

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