Sketching and Reading Quadratic Graphs from Factorised Form

The graph crosses the x-axis where \( y = 0 \). Setting \( (x \; – \; 4)(x + 1) = 0 \), either \( x \; – \; 4 = 0 \) giving \( x = 4 \), or \( x + 1 = 0 \) giving \( x = -1 \). The key step is solving each bracket equal to zero — not just reading off the numbers in the brackets.

A common error is reading the numbers directly and saying the roots are \( x = -4 \) and \( x = 1 \) — effectively getting the signs the wrong way round. This is the “same sign” misconception — students take the number they see in the bracket rather than solving the equation. A quick check: substituting \( x = 4 \) gives \( (0)(5) = 0 \) ✓, but substituting \( x = -4 \) gives \( (-8)(-3) = 24 \neq 0 \) ✗.

The y-intercept occurs where \( x = 0 \). Substituting: \( y = (0 \; – \; 5)(0 + 3) = (-5)(3) = -15 \). So the graph passes through \( (0, \; -15) \).

Students often struggle to find the y-intercept from factorised form because they are used to reading it from the expanded form \( y = ax^2 + bx + c \) where it is simply \( c \). This is the “y-intercept is only visible in expanded form” misconception. In factorised form, you can always find the y-intercept by substituting \( x = 0 \) and multiplying the constant terms: here \( (-5)(3) = -15 \). Another approach: expanding gives \( x^2 \; – \; 2x \; – \; 15 \), confirming \( c = -15 \).

The roots of \( y = (x \; – \; 6)(x \; – \; 2) \) are \( x = 6 \) and \( x = 2 \). Because a parabola is symmetric, the line of symmetry passes exactly halfway between the roots. The midpoint is \( \frac{6 + 2}{2} = 4 \), so the turning point has \( x = 4 \).

Students who haven’t connected symmetry to the turning point may try to complete the square or expand and use \( x = \frac{-b}{2a} \) without realising the factorised form gives the answer more directly. This targets the “no link between roots and vertex” misconception. Note: the full turning point is \( (4, \; -4) \) since \( y = (4 \; – \; 6)(4 \; – \; 2) = (-2)(2) = -4 \).

Expanding: \( y = -(x^2 \; – \; 6x + 5) = -x^2 + 6x \; – \; 5 \). The coefficient of \( x^2 \) is \( -1 \), which is negative. When the coefficient of \( x^2 \) is negative, the parabola opens downward — it has a maximum point rather than a minimum.

This targets the “ignoring the negative coefficient” misconception. Students frequently sketch all quadratics as U-shaped (opening upward) because they focus on the brackets and forget the leading negative sign. The negative sign means the parabola is reflected in the x-axis. For \( y = -(x \; – \; 1)(x \; – \; 5) \), the graph still crosses at \( x = 1 \) and \( x = 5 \), but it goes up from the roots to a maximum at \( x = 3 \), \( y = -(3 \; – \; 1)(3 \; – \; 5) = -(-4) = 4 \), then back down — an ∩-shape, not a ∪-shape.

The x-intercepts are found when \( y = 0 \). If we set \( (2x \; – \; 3)(x + 4) = 0 \), then either the first bracket is zero or the second bracket is zero. Solving \( 2x \; – \; 3 = 0 \) gives \( 2x = 3 \Rightarrow x = 1.5 \). Solving \( x + 4 = 0 \) gives \( x = -4 \).

This targets the common “just flip the sign” algorithmic misconception. Students frequently assume the root for \( (2x \; – \; 3) \) must simply be \( +3 \) because they are used to monic quadratics where \( x = c \) is the root for \( (x \; – \; c) \). Constructing this argument forces them to realise they must actually solve the mini-equation \( 2x \; – \; 3 = 0 \).

Example: \( y = (x + 3)(x \; – \; 7) \)

Another: \( y = 2(x + 3)(x \; – \; 7) \)

Creative: \( y = -5(x + 3)(x \; – \; 7) \) — any non-zero scalar multiple preserves the roots. The negative coefficient flips the parabola but doesn’t change where it crosses the x-axis.

Trap: \( y = (x \; – \; 3)(x + 7) \) — this has roots at \( x = 3 \) and \( x = -7 \), not \( x = -3 \) and \( x = 7 \). Students fall for this because of the “same sign” misconception: they match the sign they see in the bracket to the root rather than solving each bracket equal to zero.

Example: \( y = x(x \; – \; 4) \)

Another: \( y = x(x + 6) \)

Creative: \( y = 3x(x \; – \; 1) \) — the scalar multiple doesn’t affect whether the graph passes through the origin. As long as \( x \) is a factor, substituting \( x = 0 \) gives \( y = 0 \).

Trap: \( y = (x \; – \; 1)(x + 1) \) — students might think this passes through the origin because the roots are symmetric about zero (\( x = 1 \) and \( x = -1 \)). But substituting \( x = 0 \) gives \( (-1)(1) = -1 \), not 0. This exploits the “symmetric roots means it passes through the origin” misconception.

Example: \( y = (x \; – \; 1)(x \; – \; 5) \) — roots at 1 and 5, midpoint \( \frac{1+5}{2} = 3 \) ✓

Another: \( y = x(x \; – \; 6) \) — roots at 0 and 6, midpoint \( \frac{0+6}{2} = 3 \) ✓

Creative: \( y = (x \; – \; 3)^2 \) — this is a “repeated root” quadratic with both roots at \( x = 3 \), so the line of symmetry is \( x = 3 \). The graph just touches the x-axis at \( (3, \; 0) \).

Trap: \( y = (x \; – \; 3)(x + 3) \) — students might think “there’s a 3 in both brackets so the line of symmetry is \( x = 3 \).” But the roots are \( x = 3 \) and \( x = -3 \), giving a midpoint of \( \frac{3 + (-3)}{2} = 0 \). The line of symmetry is \( x = 0 \), not \( x = 3 \). This exploits the “reading the line of symmetry from one bracket” misconception.

Example: \( y = (x \; – \; 1)(x \; – \; 3) \) — turning point at \( x = 2 \), \( y = (2 \; – \; 1)(2 \; – \; 3) = (1)(-1) = -1 \). Turning point \( (2, \; -1) \) ✓

Another: \( y = (x + 2)(x \; – \; 4) \) — turning point at \( x = 1 \), \( y = (3)(-3) = -9 \). Turning point \( (1, \; -9) \) ✓

Creative: \( y = (x \; – \; 100)(x \; – \; 102) \) — roots at 100 and 102, turning point at \( x = 101 \), \( y = (1)(-1) = -1 \). Shows this works for any pair of distinct roots with a positive leading coefficient.

Trap: \( y = -(x \; – \; 2)(x \; – \; 6) \) — the turning point is at \( x = 4 \), \( y = -(4 \; – \; 2)(4 \; – \; 6) = -(2)(-2) = 4 \). The turning point is \( (4, \; 4) \), which is above the x-axis. Students who ignore the negative coefficient fall for the “all parabolas have a minimum below the x-axis” misconception.

Example: \( y = (x \; – \; 4)(x \; – \; 4) \) — touching the x-axis at \( x = 4 \) ✓

Another: \( y = (x + 7)^2 \) ✓

Creative: \( y = -3(x \; – \; 1)^2 \) — an upside-down parabola that just grazes the axis at \( x = 1 \).

Trap: \( y = x(x \; – \; 3) \) — students often think the lone “x” on the outside doesn’t count as a root, assuming the only intercept is at \( x = 3 \). In reality, setting \( x = 0 \) gives the second root, so this graph actually crosses at both \( 0 \) and \( 3 \).

The y-intercept is found by substituting \( x = 0 \): \( y = (0 \; – \; a)(0 \; – \; b) = (-a)(-b) = ab \). So the y-intercept equals the product \( ab \). This is negative when \( a \) and \( b \) have opposite signs (one root positive, one negative). For example, \( y = (x \; – \; 3)(x + 2) \) has y-intercept \( (-3)(2) = -6 \).

But if both roots are positive, the product \( ab \) is positive: \( y = (x \; – \; 2)(x \; – \; 5) \) has y-intercept \( (-2)(-5) = 10 \). Students who mostly see examples with one positive and one negative root develop the “y-intercept of a factorised quadratic is always negative” misconception.

The roots are \( x = a \) and \( x = b \). A parabola is always symmetric, and the axis of symmetry passes through the vertex. Since the two roots are equidistant from the axis of symmetry, it must pass through their midpoint: \( x = \frac{a + b}{2} \).

This can be verified algebraically: expanding \( y = (x \; – \; a)(x \; – \; b) = x^2 \; – \; (a+b)x + ab \), the axis of symmetry is \( x = \frac{a+b}{2} \) using \( x = \frac{-B}{2A} \). This is always true regardless of the values of \( a \) and \( b \). Students who haven’t connected the link between roots and symmetry may think this is only sometimes true.

A quadratic in factorised form can be written as \( y = k(x \; – \; a)(x \; – \; b) \) where \( k \neq 0 \). If \( a \neq b \), the graph crosses at two distinct points \( x = a \) and \( x = b \). For example, \( y = (x \; – \; 1)(x \; – \; 5) \) crosses at \( x = 1 \) and \( x = 5 \).

However, \( y = (x \; – \; 3)(x \; – \; 3) = (x \; – \; 3)^2 \) touches the x-axis only at \( x = 3 \) — one repeated root, not two distinct points. This targets the “factorised always means two distinct roots” misconception. Students forget that a repeated factor like \( (x \; – \; 3)^2 \) is still “factorised form” but produces a graph that touches rather than crosses the axis.

The turning point occurs at \( x = \frac{a+b}{2} \). Substituting into \( y = (x \; – \; a)(x \; – \; b) \) gives \( y = \left(\frac{b \; – \; a}{2}\right)\left(\frac{a \; – \; b}{2}\right) = -\frac{(a \; – \; b)^2}{4} \). Since \( a \neq b \), \( (a \; – \; b)^2 > 0 \), so \( y < 0 \). The turning point y-coordinate is always negative.

For example: \( y = (x \; – \; 1)(x \; – \; 5) \) has turning point y-value \( -\frac{(1 \; – \; 5)^2}{4} = -\frac{16}{4} = -4 \). This targets the “turning point could be anywhere” misconception — students may not realise that for a positive leading coefficient with two distinct roots, the turning point is always below the x-axis. The parabola must dip down between its roots.

The student has made the “read the numbers directly” misconception — they are taking the values and signs visible in the brackets rather than solving each bracket equal to zero. To find the roots, set each factor to zero: \( x + 3 = 0 \) gives \( x = -3 \), and \( x \; – \; 7 = 0 \) gives \( x = 7 \). The student has the signs exactly backwards.

The correct roots are \( x = -3 \) and \( x = 7 \). A useful check: substitute back. \( (-3 + 3)(-3 \; – \; 7) = (0)(-10) = 0 \) ✓, and \( (7 + 3)(7 \; – \; 7) = (10)(0) = 0 \) ✓.

The student correctly found the x-coordinate of the turning point (the line of symmetry is at \( x = 4 \)). However, they evaluated the function at \( x = 0 \) instead of \( x = 4 \) to find the y-coordinate.

Evaluating at \( x = 0 \) finds the y-intercept, not the turning point. This demonstrates the “confusing intercepts with vertices” misconception. To get the correct y-coordinate, they must substitute their x-coordinate back into the equation: \( y = (4 \; – \; 3)(4 \; – \; 5) = (1)(-1) = -1 \). The correct turning point is \( (4, -1) \).

The student has confused the x-intercepts (roots) with the y-intercept. The values \( x = 4 \) and \( x = 7 \) are where the graph crosses the x-axis — these are the roots. The y-intercept is a different feature entirely: it is where the graph crosses the y-axis, found by substituting \( x = 0 \).

Here, \( y = (0 \; – \; 4)(0 \; – \; 7) = (-4)(-7) = 28 \), so the y-intercept is \( (0, \; 28) \). This confusion often stems from students learning “the intercepts come from the brackets” without distinguishing between x-intercepts and y-intercepts. The brackets directly give x-intercepts (roots); the y-intercept requires substituting \( x = 0 \).

The student has correctly identified the roots but drawn a straight line instead of a curve — the “linear not quadratic” misconception. A quadratic expression produces a parabola (curved U-shape or ∩-shape), not a straight line.

The graph of \( y = (x \; – \; 1)(x \; – \; 5) \) is a U-shaped parabola that crosses the x-axis at \( x = 1 \) and \( x = 5 \), has a turning point at \( (3, \; -4) \), and passes through the y-intercept at \( (0, \; 5) \). Drawing a line through two roots ignores the quadratic nature of the expression.