P1 1.3, P1 1.5 - Quadratic graphs and completed squares | Edexcel A-Level International Mathematics

Read the key features of a quadratic graph and rewrite its equation to reveal its turning point. P1 1.3 connects the algebra to the graph; P1 1.5 supplies completing the square, which can reveal a minimum or maximum without plotting a table of values.

Roots and intercepts

What is the smallest possible value of $2x^2-8x+5$ ? Its expanded form hides the answer. Before uncovering it, connect the forms you already know to points on a graph.

Quadratic function

A quadratic function has the form $f(x)=ax^2+bx+c$ , where $a,b,c$ are real and $a\ne0$ . Its graph is a parabola, opening upwards if $a>0$ and downwards if $a<0$ .

The $y$ -intercept occurs at $x=0$ , so its coordinates are $(0,c)$ . A root is a value of $x$ for which $f(x)=0$ ; a real root gives an $x$ -intercept $(x,0)$ . In factorised form, a product is zero exactly when at least one factor is zero. Thus $(x-2)(x+4)=0$ gives $x=2$ or $x=-4$ .

Worked example 1: intercepts of an upward parabola

Find the axis intercepts and direction of opening of $y=(x-1)(x-5)$ .

Plan. Set each factor to zero for the $x$ -intercepts, then substitute $x=0$ .

\begin{aligned} x-1=0\text{ or }x-5=0&\quad\Rightarrow\quad x=1\text{ or }5\\ y(0)&=(-1)(-5)=5. \end{aligned}

Expanding would give leading term $x^2$ , with positive coefficient.

Answer. $(1,0)$ , $(5,0)$ and $(0,5)$ ; the parabola opens upwards.

Check. Substituting $x=1$ or $5$ makes one factor zero. At $x=3$ , $y=-4$ , so the curve does lie below the axis between its two crossings.

Change only the overall sign of the function. Multiplying by $-1$ changes nonzero heights, but zero remains zero.

Worked example 2: changing the leading sign

Find the axis intercepts and direction of opening of $y=-(x-1)(x-5)$ .

Plan. Keep the same zero factors, then include the minus sign in the $y$ -intercept.

y=0\Rightarrow x=1\text{ or }5,\qquad y(0)=-(-1)(-5)=-5.

The leading term is now $-x^2$ .

Answer. $(1,0)$ , $(5,0)$ and $(0,-5)$ ; the parabola opens downwards.

Check. At $x=3$ , $y=4$ , the opposite height from Example 1. Both original root substitutions still give zero.

The graphs show what changed and what stayed fixed.

[DIAGRAM: asset_name: Quadratic intercepts and leading sign; asset_slug: quadratic-intercepts; caption: Opposite leading signs reverse the heights but preserve the two roots.; description: The graphs y equals (x minus 1)(x minus 5) and y equals negative (x minus 1)(x minus 5) on the same labelled axes. Both cross at (1,0) and (5,0). Their y-intercepts are (0,5) and (0,negative 5).]

A root is one coordinate value; an intercept is a point.

Watch out

The root from $x+4=0$ is $-4$ , not $4$ . Report an intercept as $(-4,0)$ when coordinates are requested.

Use the factorised form directly.

Completing a monic square

Completing the square rewrites a quadratic as a squared bracket plus a constant; it does not change the function.

Expanding $(x+p)^2$ gives $x^2+2px+p^2$ . To match $x^2+6x$ , choose $p=3$ . The bracket then contains an extra $9$ , which must be subtracted outside:

x^2+6x=(x+3)^2-9.

The coefficient $1$ on $x^2$ makes this a monic quadratic.

Completing a monic square

For real $b,c$ ,

x^2+bx+c=\left(x+\frac b2\right)^2+c-\frac{b^2}{4}.

Halve the linear coefficient for the bracket; subtract the square you have added.

Not in the formula booklet: learn it.

The correction outside the bracket keeps the equality true for every $x$ .

Worked example 3: a positive linear coefficient

Complete the square for $x^2+6x+2$ .

Plan. Use $(x+3)^2$ , then adjust its constant from $9$ to $2$ .

\begin{aligned} x^2+6x+2&=(x+3)^2-9+2 &&\text{compensate for the added }9\\ &=(x+3)^2-7. \end{aligned}

Answer. $(x+3)^2-7$ .

Check. Expanding gives $x^2+6x+9-7=x^2+6x+2$ .

Change only the sign of the linear term. The bracket sign changes; the square of its constant does not.

Worked example 4: a negative linear coefficient

Complete the square for $x^2-6x+2$ .

Plan. Halve $-6$ to get $-3$ , then subtract the extra $9$ .

x^2-6x+2=(x-3)^2-9+2=(x-3)^2-7.

Answer. $(x-3)^2-7$ .

Check. At $x=1$ , both forms give $-3$ . Expanding the square gives the required middle term $-6x$ .

The bracket constant is half the linear coefficient, not half the constant term.

Watch out

$x^2+6x+2\ne(x+3)^2+2$ . The right side has constant $11$ . Subtracting the added $9$ repairs the equality.

Keep the correction outside the square.

Completing a non-monic square

When the coefficient of $x^2$ is not $1$ , take it outside the quadratic and linear terms before completing the square.

For $2x^2+8x+1$ , start with $2(x^2+4x)+1$ . Completing the square inside gives $2[(x+2)^2-4]+1$ . The outside $2$ multiplies the correction as well as the square.

Completing a general quadratic

For real $a,b,c$ , with $a\ne0$ ,

ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.

Factoring $a$ out first explains each term; the formula also works when $a<0$ .

Not in the formula booklet: learn it.

Keep the outside coefficient visible until the final constant is simplified.

Worked example 5: an outside factor of two

Complete the square for $2x^2-8x+5$ .

Plan. Factor $2$ from the first two terms, then complete the square inside.

\begin{aligned} 2x^2-8x+5&=2(x^2-4x)+5 &&\text{make the inside monic}\\ &=2[(x-2)^2-4]+5 &&\text{complete the inside square}\\ &=2(x-2)^2-3 &&\text{apply }2\text{ to the correction} \end{aligned}

Answer. $2(x-2)^2-3$ .

Check. Re-expanding gives $2x^2-8x+8-3=2x^2-8x+5$ . At $x=0$ , both forms give $5$ .

Change only the leading coefficient in the original expression from $2$ to $-2$ . The inside linear coefficient changes when that factor is removed.

Worked example 6: a negative outside factor

Complete the square for $-2x^2-8x+5$ .

Plan. Factor $-2$ from the first two terms, so the inside square starts with $x^2+4x$ .

\begin{aligned} -2x^2-8x+5&=-2(x^2+4x)+5\\ &=-2[(x+2)^2-4]+5\\ &=-2(x+2)^2+13. \end{aligned}

The correction becomes positive because $(-2)(-4)=8$ .

Answer. $-2(x+2)^2+13$ .

Check. At $x=-2$ , the original is $-8+16+5=13$ , matching the completed form. Expanding also restores all three coefficients.

The correction belongs inside the bracket until multiplied out.

Watch out

$2[(x-2)^2-4]+5$ becomes $2(x-2)^2-8+5$ , not $2(x-2)^2-4+5$ . The outside factor multiplies both terms.

Use the same method with a different coefficient.

Vertex and symmetry from completed form

In $y=a(x-h)^2+k$ , the squared term is zero at $x=h$ , and equal positive distances on either side of $h$ give equal heights.

Vertex and axis of symmetry

The vertex, or turning point, of $y=a(x-h)^2+k$ is $(h,k)$ . The axis of symmetry is the vertical line $x=h$ . If $a>0$ , the vertex is a minimum; if $a<0$ , it is a maximum.

The reason is that $(x-h)^2\ge0$ for real $x$ . Multiplying by positive $a$ keeps it nonnegative, so $y\ge k$ . Multiplying by negative $a$ makes it nonpositive, so $y\le k$ . There is no real value of $x$ that can push the graph past that bound.

Worked example 7: finding a minimum

Find the vertex, axis of symmetry and minimum value of $y=2x^2-8x+5$ .

Plan. Use Example 5's completed form and the nonnegative square.

y=2(x-2)^2-3\ge-3,

with equality exactly when $x-2=0$ .

Answer. Vertex $(2,-3)$ , axis $x=2$ , minimum value $-3$ at $x=2$ . This answers the opening question.

Check. At $x=1$ and $x=3$ , the original function gives $-1$ in each case, consistent with symmetry and both above the minimum.

Change only the coefficient of the squared term from $2$ to $-2$ , keeping the same vertex.

Worked example 8: finding a maximum

Find the vertex, axis of symmetry and maximum value of $y=-2(x-2)^2-3$ .

Plan. A negative multiple of a nonnegative square cannot exceed zero.

-2(x-2)^2\le0\quad\Rightarrow\quad y\le-3,

with equality at $x=2$ .

Answer. Vertex $(2,-3)$ , axis $x=2$ , maximum value $-3$ at $x=2$ .

Check. At $x=1$ and $3$ , $y=-5$ , equal heights below $-3$ . At the vertex, substitution gives $y=-3$ .

Both curves have the same turning point, while the sign determines which way they open.

[DIAGRAM: asset_name: A square reveals a turning point; asset_slug: completed-square-vertices; caption: The same squared bracket gives the same symmetry line; its coefficient decides minimum or maximum.; description: Two side-by-side graphs, y equals 2(x minus 2) squared minus 3 and y equals negative 2(x minus 2) squared minus 3. Each has vertex (2,negative 3) and dashed symmetry line x equals 2; the first has a minimum and the second a maximum.]
The same squared bracket gives the same symmetry line

Read the value that makes the whole bracket zero.

Watch out

$(x+4)^2=(x-(-4))^2$ , so $y=3(x+4)^2-1$ has vertex $(-4,-1)$ , not $(4,-1)$ . A minimum value is a height; its location is an $x$ -value.

Combine completing the square with a turning-point statement.

Quick Recap

Factorised form exposes real roots; $x=0$ gives the $y$ -intercept.
Complete a monic square by halving the linear coefficient and correcting the constant.
With a non-monic quadratic, apply the outside coefficient to the correction too.
$a(x-h)^2+k$ has vertex $(h,k)$ , symmetry line $x=h$ , and minimum or maximum height $k$ .

The smallest value of $2x^2-8x+5$ is $-3$ , reached at $x=2$ .

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P1 1.3, P1 1.5 - Quadratic graphs and completed squares

Roots and intercepts

Completing a monic square

Completing a non-monic square

Vertex and symmetry from completed form

Quick Recap

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