6.03d, 7.01c - Turning points and intersections | OCR GCSE Mathematics

Can an equation reveal a curve's lowest point before you draw it? Use completed-square form to identify turning points, then connect curve–line intersections with simultaneous equations. This lesson is for Higher tier.

The turning point in completed-square form

For $y=x^2+6x+7$ , which input gives the smallest possible output? A table might suggest an answer; completing the square explains why it is the minimum.

You already know how to complete a square. Here,

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

The square is never negative, and reaches zero precisely at $x=-3$ . The smallest height is therefore $-2$ .

Turning point from a completed square

For $y=a(x-h)^2+k$ , with $a\ne0$ , the turning point is $(h,k)$ and the symmetry line is $x=h$ . It is a minimum if $a>0$ , or a maximum if $a<0$ . Not on either 2027 formula sheet: learn how the square explains it.

Read the input that makes the bracket zero, rather than copying the sign printed inside it. The bracket $x+3$ is zero at $-3$ .

Worked example 1: the minimum hidden in an expansion

Question. Find the turning point of $y=x^2+6x+7$ , stating its type and symmetry line.

Plan. Complete the square and use the smallest possible value of a square.

\begin{aligned} y&=(x+3)^2-9+7 &&\text{preserve the original expression}\\ &=(x+3)^2-2\\ (x+3)^2&\ge0 &&\text{every real square is non-negative}\\ y&\ge-2 &&\text{equality when }x=-3 \end{aligned}

Answer. Minimum $(-3,-2)$ , with symmetry line $x=-3$ .

Check. Substituting $x=-3$ in the original gives $9-18+7=-2$ . At $-4$ and $-2$ , the height is $-1$ in both cases, confirming the equal-height pattern around the turning point.

Negate the whole quadratic. This reverses every height, turning a minimum into a maximum without moving the symmetry line.

Worked example 2: a maximum instead

Question. Find the turning point of $y=-x^2-6x-7$ , stating its type and symmetry line.

Plan. Factor out the negative sign, then use the completed square from the corresponding positive expression.

\begin{aligned} y&=-(x^2+6x+7)\\ &=-\big((x+3)^2-2\big)\\ &=-(x+3)^2+2\\ y&\le2 &&\text{negative square is at most zero} \end{aligned}

Answer. Maximum $(-3,2)$ , with symmetry line $x=-3$ .

Check. The original expression gives $-9+18-7=2$ at $-3$ . At $-4$ and $-2$ , it gives $1$ , below that maximum.

[DIAGRAM: asset_name: A completed square locates the turning point; asset_slug: completed-square-turns; caption: The bracket is zero at the symmetry line and the sign decides minimum or maximum.; description: Two scaled graphs show x squared plus six x plus seven and its negative. The minimum negative three negative two and maximum negative three two are labelled, and each has symmetry line x equals negative three.]
The bracket is zero at the symmetry line and the sign decides minimum or maximum.

Watch out

In $y=(x+3)^2-2$ , the turning point is $(-3,-2)$ , not $(3,-2)$ . Solve $x+3=0$ to find the input.

Use the square to justify the type as well as locating the point.

Intersections of a curve and line

A line and a curve share a point only when its coordinates satisfy both equations, just as with two straight lines.

At an intersection of $y=x^2-2x-3$ and $y=x-3$ , their heights are equal, so $x^2-2x-3=x-3$ . Solving that equation gives the horizontal positions. Substitute each one into either original equation to find its matching height. Keep each $x$ paired with its own $y$ .

A graph can estimate these coordinates; algebra can often give exact values. Two crossings give two solution pairs. A line touching a quadratic at its turning point can give one pair, while a line missing the curve gives no real solution. For example, $y=0$ meets $y=x^2$ only at $(0,0)$ , whereas $y=-1$ never meets it because a square cannot be negative.

Worked example 3: two exact intersection points

Question. Find the intersections of $y=x^2-2x-3$ and $y=x-3$ .

Plan. Equate the heights, solve the resulting quadratic, then calculate each corresponding $y$ .

\begin{aligned} x^2-2x-3&=x-3 &&\text{equal heights at an intersection}\\ x^2-3x&=0 &&\text{collect on one side}\\ x(x-3)&=0\\ x&=0\text{ or }3\\ x=0&\Rightarrow y=-3 &&\text{use the line}\\ x=3&\Rightarrow y=0 \end{aligned}

Answer. Intersections $(0,-3)$ and $(3,0)$ , giving two simultaneous solution pairs.

Check. The curve gives $-3$ at $0$ , and $9-6-3=0$ at $3$ . Both pairs also satisfy the line.

Raise only the line's intercept from $-3$ to $1$ . The curve stays fixed, but both intersections move.

Worked example 4: pairing the two new solutions

Question. Find the intersections of $y=x^2-2x-3$ and $y=x+1$ .

Plan. Equate expressions, factorise, and calculate the matching line height for each root.

\begin{aligned} x^2-2x-3&=x+1\\ x^2-3x-4&=0\\ (x-4)(x+1)&=0\\ x&=4\text{ or }-1\\ x=4&\Rightarrow y=5\\ x=-1&\Rightarrow y=0 \end{aligned}

Answer. $(-1,0)$ and $(4,5)$ .

Check. The curve heights are $1+2-3=0$ and $16-8-3=5$ , independently confirming the two values found from the line.

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6.03d, 7.01c - Turning points and intersections

The turning point in completed-square form

Intersections of a curve and line

Continue this lesson.

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