The parabola's mirror line
Tonight's idea
The graph of a quadratic such as \(y = (x - 1)(x - 5)\) is a U-shaped curve called a parabola. A product is zero only when a factor is, so \(y = 0\) at \(x = 1\) and \(x = 5\). Those are the places it crosses the \(x\)-axis.
Half way between the crossings is \(x = 3\). Step a distance \(d\) to either side and work out the height:
\[\begin{aligned} x = 3 + d:&\quad y = (2 + d)(-2 + d) = d^2 - 4 \\ x = 3 - d:&\quad y = (2 - d)(-2 - d) = d^2 - 4 \end{aligned}\]
The same height, whatever \(d\) is. So the line \(x = 3\) is a mirror: the left half of the curve is the reflection of the right half. The turning point, called the vertex, has no partner, so it must lie on the mirror. Put \(d = 0\): the vertex is \((3, -4)\). And since \(d^2\) is never negative, \(y\) never goes below \(-4\).
The mirror line is half way between any two points at the same height. The vertex is on it.
Watch one: the lowest point of \(y = x^2 - 2x - 8\)
Factor: \(y = (x + 2)(x - 4)\). The curve crosses the axis at \(x = -2\) and \(x = 4\).
The mirror line is half way between: \(x = \dfrac{-2 + 4}{2} = 1\).
The vertex is on it. At \(x = 1\), \(y = 1 - 2 - 8 = -9\). The lowest point is \((1, -9)\).
Check the symmetry one step either side: \(x = 0\) and \(x = 2\) both give \(y = -8\).
Drag a parabola by its vertex and watch the crossings move with it: mathdetours.com/demos/quadratic.html. Next year you will find the vertex of any quadratic, factorable or not.
Your turn
- For \(y = (x - 2)(x - 8)\), find (a) where it crosses the \(x\)-axis, (b) the mirror line, (c) the vertex.
- Factor \(y = x^2 + 6x + 5\). Then find its crossings, mirror line and vertex. Sketch it.
- A ball is thrown straight up. After \(t\) seconds its height is \(h = 5t(4 - t)\) metres. (a) When does it land? (b) When is it highest, and how high is that? (c) It passes 15 m on the way up at \(t = 1\). When does it pass 15 m on the way down?
- \(y = x^2 - 6x + 20\) never crosses the \(x\)-axis, so there are no crossings to use. (a) Find the two values of \(x\) where \(y = 20\). (b) Use them to find the mirror line and the vertex. (c) How does the vertex show the curve never reaches the axis?
Stretch: the biggest pen
You have 40 m of fence to make three sides of a rectangular pen. A long wall is the fourth side. If the two sides leaving the wall are \(x\) metres each, the area is \(A = x(40 - 2x)\). Find the largest possible area and the pen's measurements. This parabola is upside down.
Answers
- (a) \(x = 2\) and \(x = 8\). (b) \(x = 5\). (c) \(y = (5 - 2)(5 - 8) = -9\), so the vertex is \((5, -9)\).
- \(y = (x + 1)(x + 5)\). Crossings at \(x = -1\) and \(x = -5\). Mirror line \(x = -3\). There \(y = 9 - 18 + 5 = -4\), so the vertex is \((-3, -4)\). The sketch is a U through \((-5, 0)\), \((-3, -4)\), \((-1, 0)\) and \((0, 5)\).
- (a) \(h = 0\) at \(t = 0\) (the throw) and \(t = 4\): it lands after 4 seconds. (b) Half way, at \(t = 2\): \(h = 5 \times 2 \times 2 = 20\) m. (c) \(t = 1\) is one second before the mirror, so the partner is one second after: \(t = 3\). Check: \(5 \times 3 \times 1 = 15\).
- (a) \(x^2 - 6x + 20 = 20\) means \(x^2 - 6x = 0\), so \(x(x - 6) = 0\): \(x = 0\) or \(x = 6\). (b) Two points at the same height, so the mirror is half way: \(x = 3\). There \(y = 9 - 18 + 20 = 11\). Vertex \((3, 11)\). (c) The vertex is the lowest point and it is 11 above the axis.
Stretch. \(A = 0\) at \(x = 0\) and at \(x = 20\). The mirror line is \(x = 10\), and the vertex is the highest point because the curve opens downward. \(A = 10 \times 20 = 200\) m². The pen is 10 m out from the wall and 20 m along it: twice as long as it is deep, not a square.