Sketching and Reading Quadratic Graphs in Completed Square Form

The completed square form $y = (x \; – \; 3)^2 + 1$ tells us the turning point directly. The squared bracket $(x \; – \; 3)^2$ is always $\ge 0$, and it equals 0 when $x = 3$ (since $3 \; – \; 3 = 0$). At that point, $y = 0 + 1 = 1$. So the minimum is at $(3, 1)$.

If the vertex were at $x = -3$, then substituting gives $y = (-3 \; – \; 3)^2 + 1 = 36 + 1 = 37$ — that’s clearly not a minimum! The sign inside the bracket is the opposite of the x-coordinate of the vertex — this is the sign reversal error, one of the most common mistakes when reading completed square form.

The y-intercept is where $x = 0$. Substituting: $y = (0 \; – \; 4)^2 \; – \; 7 = 16 \; – \; 7 = 9$. So the y-intercept is $(0, 9)$, not $(0, -7)$.

The value $-7$ is the minimum value of the graph (the y-coordinate of the turning point), not where it crosses the y-axis. Students often make the confusing the minimum value with the y-intercept error because the $-7$ is the “loose number” they can see at the end of the expression. To find the y-intercept, you must always substitute $x = 0$ — the squared bracket doesn’t vanish, it contributes $4^2 = 16$.

In completed square form, the sign in front of the squared bracket determines the shape. Here, $-(x + 2)^2$ is always $\le 0$ (a non-negative square multiplied by $-1$). The largest value this can take is 0, which happens when $x = -2$. At that point, $y = 0 + 5 = 5$. For any other value of $x$, we subtract something positive from 5, making $y \lt 5$.

This means the graph is an upside-down parabola ($\cap$-shape) with highest point $(-2, 5)$. The “all quadratics have a minimum” misconception comes from only encountering positive leading coefficients. A negative coefficient flips the parabola so the vertex becomes a maximum.

The line of symmetry of a parabola always passes through the turning point and is a vertical line. The turning point of $y = (x + 6)^2 \; – \; 10$ is at $(-6, -10)$, so the line of symmetry is $x = -6$. Students who say $x = -10$ are making the vertex coordinate confusion — mixing up the x-coordinate and y-coordinate of the turning point. The $-10$ is the minimum value (the y-coordinate), not the location along the x-axis.

Think of the squared bracket $(x + 6)^2$ as a distance measuring tool. Because squaring makes everything positive (or zero), the lowest possible value is achieved when this “distance” is exactly zero — i.e., when $x = -6$.

A quick check: the graph is symmetric about $x = -6$, so plugging in $x = -5$ and $x = -7$ should give the same y-value. Both give $y = 1 \; – \; 10 = -9$. ✓ If the axis were $x = -10$, then $x = -9$ and $x = -11$ would need to give the same y-value: $(-9 + 6)^2 \; – \; 10 = 9 \; – \; 10 = -1$ and $(-11 + 6)^2 \; – \; 10 = 25 \; – \; 10 = 15$. They’re different, so $x = -10$ is not the axis of symmetry.

Example: $y = (x \; – \; 1)^2 \; – \; 4$ — turning point at $(1, -4)$, which is below the x-axis since $-4 \lt 0$.

Another: $y = (x + 3)^2 \; – \; 1$ — turning point at $(-3, -1)$, below the x-axis.

Creative: $y = (x \; – \; 100)^2 \; – \; 0.001$ — the turning point is only just below the x-axis at $(100, -0.001)$, showing it doesn’t need to be far below.

Trap: $y = (x \; – \; 4)^2 + 2$ — the +2 might look small, but the turning point is at $(4, 2)$, which is above the x-axis. A turning point is below the x-axis only when the $k$ value is negative.

Example: $y = (x \; – \; 1)^2 + 3$ — minimum at $(1, 3)$, which is above the x-axis, so the graph never reaches $y = 0$.

Another: $y = (x + 5)^2 + 1$ — minimum at $(-5, 1)$, entirely above the x-axis.

Creative: $y = -(x \; – \; 2)^2 \; – \; 6$ — this is an upside-down parabola with maximum at $(2, -6)$. Since the highest point is below the x-axis, it never crosses it either. This catches students who only think of U-shaped graphs.

Trap: $y = (x \; – \; 5)^2$ — this looks like it might not cross the x-axis because there’s no “−” or “+” at the end, but the turning point is at $(5, 0)$, which means the graph touches the x-axis. Touching gives a repeated root, so this quadratic does have a real root and is a non-example.

Example: $y = (x \; – \; 2)^2 + 1$ and $y = 3(x \; – \; 2)^2 + 1$ — both have turning point at $(2, 1)$, but the second is narrower (stretched vertically by a factor of 3).

Another: $y = (x \; – \; 2)^2 + 1$ and $y = -(x \; – \; 2)^2 + 1$ — same turning point $(2, 1)$, but one opens upward and the other opens downward.

Creative: $y = \frac{1}{2}(x \; – \; 2)^2 + 1$ and $y = 5(x \; – \; 2)^2 + 1$ — same vertex, but one is very wide and the other very narrow. The coefficient only affects the “steepness,” not the vertex location.

Trap: $y = (x \; – \; 2)^2 + 1$ and $y = (x \; – \; 2)^2 + 3$ — these might look different, but the shape is identical — they’re just translated vertically. Sliding a graph up the wall doesn’t change how wide it is! The turning points are different: $(2, 1)$ vs $(2, 3)$. To change the shape you need a different coefficient in front of the squared bracket. This is the vertical translation changes shape misconception.

Example: $y = (x + 3)^2 \; – \; 4$ — expanding: $x^2 + 6x + 9 \; – \; 4 = x^2 + 6x + 5$ ✓

Another: We can verify via factorised form: $(x + 3)^2 \; – \; 4 = (x + 3 \; – \; 2)(x + 3 + 2) = (x + 1)(x + 5)$. Setting $x^2 + 6x + 5 = 0$ gives $(x + 1)(x + 5) = 0$, so roots at $x = -1$ and $x = -5$. Both forms produce the same roots — confirming the equivalence.

Creative: $y = \frac{1}{4}(2x + 6)^2 \; – \; 4$ — an unusual completed square form, but expanding confirms: $\frac{1}{4}(4x^2 + 24x + 36) \; – \; 4 = x^2 + 6x + 9 \; – \; 4 = x^2 + 6x + 5$ ✓

Trap: $y = (x + 3)^2 + 4$ — a student might halve the 6 to get +3 (correct!) but then miscalculate the adjustment, writing +4 instead of −4. Expanding gives $x^2 + 6x + 9 + 4 = x^2 + 6x + 13$, which is not the same quadratic.

Example: $y = 1.25(x \; – \; 2)^2 \; – \; 5$ — turning point is at $(2, -5)$. Substituting $x = 0$ gives $1.25(-2)^2 \; – \; 5 = 1.25(4) \; – \; 5 = 5 \; – \; 5 = 0$. ✓

Another: $y = \frac{5}{4}(x \; – \; 2)^2 \; – \; 5$ — the exact same function but written with fractions, which is often easier for students to verify without a calculator.

Creative: $y = \frac{5}{4}(x \; – \; 2)^2 \; – \; \frac{20}{4}$ — avoiding the mixed format of fractions and integers by writing the translation as an unsimplified fraction.

Trap: $y = (x \; – \; 2)^2 \; – \; 5$ — this has the correct turning point at $(2, -5)$, but students might forget to solve for the stretch factor $a$. If we check the origin: $(0 \; – \; 2)^2 \; – \; 5 = 4 \; – \; 5 = -1 \neq 0$. This graph misses the origin entirely.

This is only true when the bracket contains just $x$ with no constant — i.e. when the form is $y = ax^2 + k$. For example, $y = x^2 + 3$ has its turning point at $(0, 3)$, which is on the y-axis. But $y = (x \; – \; 4)^2 + 1$ has its turning point at $(4, 1)$, which is not on the y-axis.

The turning point is on the y-axis only when $h = 0$ in $y = a(x \; – \; h)^2 + k$. Students who think this is always true may have only seen examples like $y = x^2 + c$ and overgeneralised from that limited experience.

The line of symmetry of a parabola always passes through its vertex, which is at $(h, k)$. Since the vertex’s x-coordinate is $h$, the vertical line $x = h$ is always the axis of symmetry. This is true regardless of the values of $a$, $h$, or $k$, and whether the parabola opens upward or downward.

Students might think this fails when $a$ is negative, but changing the sign of $a$ flips the parabola vertically — it doesn’t move it sideways. The axis of symmetry is determined only by what’s inside the squared bracket. For example, $y = -(x \; – \; 3)^2 + 5$ still has its axis of symmetry at $x = 3$.

A quadratic can cross the x-axis in 0, 1, or 2 places depending on the position of the vertex relative to the x-axis. For an upward parabola $y = (x \; – \; h)^2 + k$: if $k \lt 0$, the vertex is below the x-axis and the graph crosses twice; if $k = 0$, it touches once; if $k \gt 0$, it doesn’t cross at all. Similar logic applies to downward parabolas with a negative coefficient.

True case (2 crossings): $y = (x \; – \; 1)^2 \; – \; 9$ — vertex at $(1, -9)$ is below the x-axis. Setting $y = 0$: $(x \; – \; 1)^2 = 9$, giving $x = 4$ or $x = -2$. False case (0 crossings): $y = (x + 2)^2 + 5$ — vertex at $(-2, 5)$ is above the x-axis, so the graph never reaches $y = 0$.

The y-coordinate of the turning point is $k$ in $y = a(x \; – \; h)^2 + k$. The y-intercept is found by setting $x = 0$: $y = a(0 \; – \; h)^2 + k = ah^2 + k$. These are equal only when $ah^2 = 0$, which (for a quadratic where $a \neq 0$) means $h = 0$ — i.e. the vertex is on the y-axis.

True case: $y = x^2 + 4$ — vertex at $(0, 4)$ and y-intercept = $0 + 4 = 4$. They match because the vertex is on the y-axis. False case: $y = (x \; – \; 3)^2 + 2$ — vertex at $(3, 2)$, but y-intercept = $9 + 2 = 11$. Very different!

Assuming $k \neq 0$, these graphs will never share the exact same roots. A common misconception is that a vertical stretch leaves roots invariant. However, the stretch by a factor of 2 applies only to the squared bracket, not the whole expression.

For the first equation, the roots occur when $(x \; – \; h)^2 = k$. For the second equation, roots occur when $2(x \; – \; h)^2 = k$, which means $(x \; – \; h)^2 = \frac{k}{2}$. Since $k \neq \frac{k}{2}$ (for non-zero $k$), the roots must be different. To have the same roots, the second equation would need to be $y = 2((x \; – \; h)^2 \; – \; k)$, which expands to $y = 2(x \; – \; h)^2 \; – \; 2k$.

The student has made the “same sign” error when reading the vertex. In completed square form $y = (x \; – \; h)^2 + k$, the vertex is at $(h, k)$. The expression $(x + 7)^2$ should be read as $(x \; – \; (-7))^2$, so $h = -7$, not $+7$. The correct turning point is $(-7, -2)$.

The trap is that $(x + 7)$ means the graph has shifted 7 units to the left, so the vertex has a negative x-coordinate. Students can check by substituting: when $x = -7$, $y = 0 \; – \; 2 = -2$ ✓. When $x = 7$, $y = 196 \; – \; 2 = 194$ — clearly not a minimum!

The answer happens to be correct — substituting $x = 0$ gives $y = 0 + 6 = 6$. But the student is relying on the “last number = y-intercept” misconception. This shortcut appears to work here because the quadratic has no linear term and the vertex is on the y-axis.

However, the same student would look at $y = (x \; – \; 3)^2 + 6$ and claim the y-intercept is 6, when it’s actually $(0 \; – \; 3)^2 + 6 = 9 + 6 = 15$. The correct method is always to substitute $x = 0$ into the equation. The student got lucky because the vertex happened to be on the y-axis, making $k$ equal to the y-intercept.

The student has made the “turning point = roots” misconception — confusing the coordinates of the turning point with the positions where the graph crosses the x-axis. The values 2 and −9 give the vertex at $(2, -9)$, not the x-intercepts.

To find where the graph crosses the x-axis, set $y = 0$: $(x \; – \; 2)^2 \; – \; 9 = 0$, so $(x \; – \; 2)^2 = 9$, giving $x \; – \; 2 = \pm 3$, so $x = 5$ or $x = -1$. The correct x-intercepts are $(5, 0)$ and $(-1, 0)$. The turning point tells you the lowest point of the graph, and you need to solve an equation to find where the graph actually crosses the x-axis.

The student has ignored the effect of the leading coefficient on the graph’s width. While the turning point $(1, 2)$ is correct, the coefficient 3 makes the parabola narrower (steeper) than $y = (x \; – \; 1)^2 + 2$. For instance, at $x = 2$: without the 3, $y = 1 + 2 = 3$; with the 3, $y = 3(1) + 2 = 5$. The graph rises three times as fast.

A sketch of $y = 3(x \; – \; 1)^2 + 2$ should show a noticeably narrower parabola than $y = (x \; – \; 1)^2 + 2$. The coefficient $a$ in $y = a(x \; – \; h)^2 + k$ controls the width (when $|a| \gt 1$, narrower; when $0 \lt |a| \lt 1$, wider) — it’s not just decoration.