A straight line first
The equation y = x says that y is always equal to x. Its points include (−3, −3), (−1, −1), (0, 0), (2, 2) and (3, 3). Each step of 1 to the right takes the line 1 up, the same amount every time, and that is why its graph is a straight line through the origin.
The points from (−3, −3) to (3, 3) lie on the straight line y = x, which rises 1 for every 1 across, all the way along.
Square every x
Now square each input instead: . For x = 0, 1, 2 and 3 the values of y are 0, 1, 4 and 9. For the negative inputs, a negative number times itself is positive, so , and . Negative inputs come back positive, and x and −x always give the same y.
The steps up are no longer equal. From x = 0 to 1, y rises by 1; from 1 to 2 it rises by 3; from 2 to 3 it rises by 5. Each step is steeper than the one before, so the points cannot lie on a straight line. Join them with one smooth curve.
The seven points of , from (−3, 9) to (3, 9), joined by one smooth curve. The straight line y = x is drawn beside it.
The vertex and the axis of symmetry
Read the curve from left to right. It falls, from 9 to 4 to 1, turns at (0, 0), then climbs through 1, 4 and 9. The point where it turns is the turning point, also called the vertex. For it is the lowest point of the graph, because a square is never negative, so y is never less than 0.
The curve is symmetric about the y-axis, because x and −x give the same value of y: fold the graph along the y-axis and the two halves fit. The line of the fold is the axis of symmetry, and it passes through the vertex. A curve of this shape is called a parabola.
The curve turns at its vertex (0, 0). The points (−2, 4) and (2, 4) are the same height, one on each side of the axis of symmetry, x = 0.
The shape of
Any equation , where a is not 0, has a parabola for its graph. The number a decides which way it opens. When a is positive, as in , the parabola opens upward and the vertex is the lowest point. When a is negative, as in , every value of y is the negative of its value for , so the parabola is reflected in the x-axis: it opens downward, and the vertex is the highest point.
The number c is the value of y at x = 0, so the parabola crosses the y-axis at (0, c). In , every value of y is 4 less than in , so the whole curve moves 4 down: the vertex is (0, −4), and the curve crosses the x-axis where , at x = −2 and x = 2.
Symmetry finds the vertex
Take . It crosses the x-axis where y = 0. Factorize: , so y = 0 at x = 1 and at x = 5. These values of x are the roots.
The axis of symmetry is halfway between the two roots: . The vertex lies on the axis, so put x = 3 into the equation: . The vertex is (3, −4), the lowest point, because a = 1 is positive.
Any two points at the same height are also symmetric about the axis. At x = 0 the curve is at y = 5, and at x = 6 it is at as well, so the axis is halfway between them too: (0 + 6) ÷ 2 = 3. When a curve does not cross the x-axis, a pair of equal heights finds the axis instead.
crosses the x-axis at 1 and 5 and is at height 5 at x = 0 and x = 6. Both pairs are centered on x = 3, where the vertex is (3, −4).
sum 2, product −3: x² − (sum)x + product, so the middle coefficient is −2 and the constant is −3; the axis of symmetry x = 1 is half the sum
Put the roots at 2 and 6 and read the expansion
The two handles are the roots of . Move either root and the parabola follows, with its axis of symmetry always at half their sum, halfway between them.
Reading a parabola in a real problem
When a parabola describes something real, such as the path of a jet of water, each feature of the graph answers a question. The roots are where the height is 0. The value of y at x = 0 is the height where the measuring starts, such as at a wall. The vertex gives the greatest or least value, and the x-coordinate of the vertex says where it happens.
Check each root against the situation. A root can be a correct solution of the equation and still describe a place the real object never reaches, such as a distance behind the wall a jet comes out of.
Worked example: A Water Jet from a Nozzle on a Pool Wall: Where It Lands and How High It Rises
Question A nozzle on the wall of a pool sends out a jet of water. At a horizontal distance of x m from the wall the jet is y m above the water, where y = −14x2 + x + 3. (a) How far from the wall does the jet land on the water? (b) Find the greatest height of the jet above the water, and the distance from the wall at which the jet reaches it.
1.The jet lands where its height is zero, so put y = 0: −14x2 + x + 3 = 0. Multiply both sides by −4 to clear the fraction and make the x2 term positive: x2 − 4x − 12 = 0.
The jet lands where its height is zero. Put y = 0 and multiply both sides by −4: x2 − 4x − 12 = 0. 2.Factorize. Two numbers that multiply to −12 and add to −4 are −6 and 2, so (x − 6)(x + 2) = 0 and x = 6 or x = −2.
Factorize: (x − 6)(x + 2) = 0, so the curve meets the x-axis at x = 6 and at x = −2. 3.(a) The root x = −2 is behind the wall, where there is no jet, so it is rejected. The jet lands 6 m from the wall. Check: −14 × 36 + 6 + 3 = −9 + 9 = 0.
(a) The root x = −2 is behind the wall, where there is no jet, so it is rejected. The jet lands 6 m from the wall. 4.The axis of symmetry is halfway between the two roots of the curve: x = −2 + 62 = 2. The coefficient of x2 is negative, so the parabola opens downward and its turning point is the highest point.
The parabola is symmetrical about the vertical line halfway between its two roots: x = −2 + 62 = 2. It opens downward, so its turning point is the highest point. 5.Put x = 2 into the equation: y = −14 × 4 + 2 + 3 = −1 + 5 = 4.
Put x = 2 into the equation: y = −14 × 4 + 2 + 3 = 4. 6.(b) The greatest height is 4 m, reached 2 m from the wall. Check the symmetry: at x = 0 and at x = 4, which are 2 m on either side of the axis, the height is 3 m both times.
(b) The greatest height is 4 m, reached 2 m from the wall. At x = 0 and at x = 4, the same distance on either side of the axis, the height is 3 m both times.
Answer: (a) 6 m from the wall; (b) 4 m above the water, reached 2 m from the wall
Common mistakes
- Taking the greatest height to be halfway along the jet, at x = 3. The jet starts 3 m above the water, not at the water, so the axis of symmetry is halfway between the two roots −2 and 6, which is x = 2.
- Reading the constant term 3 as the greatest height. The constant term is the height at x = 0, which is the height of the nozzle. The jet rises after it leaves the nozzle.
More quadratic graphs and coordinate geometry problems, worked step by step →
Worked example: The Main Cable of a Footbridge: Its Lowest Point and the Length of a Hanger
Question The main cable of a footbridge hangs between the tops of two towers. At a horizontal distance of x m from the left tower the cable is y m above the deck, where y = 18x2 − 2x + 10. The top of each tower is 10 m above the deck. (a) Find the distance between the towers, and the height of the lowest point of the cable above the deck. (b) A vertical hanger joins the cable to the deck 2 m from the left tower. Find the length of this hanger, and the position of the other hanger that has the same length.
1.The cable meets the tower tops where y = 10: 18x2 − 2x + 10 = 10, so 18x2 − 2x = 0. Multiply both sides by 8: x2 − 16x = 0.
The cable meets the tower tops where y = 10, so 18x2 − 2x = 0. Multiply both sides by 8: x2 − 16x = 0. 2.Factorize: x(x − 16) = 0, so x = 0 or x = 16. The left tower is at x = 0 and the right tower is at x = 16, so the towers are 16 m apart.
Factorize: x(x − 16) = 0, so x = 0 or x = 16. The towers are 16 m apart. 3.The axis of symmetry is halfway between the towers, at x = 8. The coefficient of x2 is positive, so the parabola opens upward and its turning point is the lowest point. Put x = 8: y = 18 × 64 − 16 + 10 = 8 − 16 + 10 = 2.
The axis of symmetry is halfway between the towers, at x = 8. The parabola opens upward, so its turning point is the lowest point: y = 18 × 64 − 16 + 10 = 2. 4.(a) The towers are 16 m apart, and the lowest point of the cable is 2 m above the deck, midway between them.
(a) The towers are 16 m apart, and the lowest point of the cable is 2 m above the deck, midway between them. 5.For the hanger, put x = 2: y = 18 × 4 − 4 + 10 = 0.5 + 6 = 6.5. The hanger is 6.5 m long.
Put x = 2: y = 18 × 4 − 4 + 10 = 6.5, so the hanger is 6.5 m long. 6.(b) The hanger at x = 2 is 6 m to the left of the axis x = 8. The point 6 m to the right of the axis is x = 14, and the cable has the same height there. Check: 18 × 196 − 28 + 10 = 24.5 − 18 = 6.5.
(b) The hanger at x = 2 is 6 m to the left of the axis. The cable has the same height 6 m to the right of the axis, at x = 14.
Answer: (a) 16 m apart, and the lowest point is 2 m above the deck; (b) 6.5 m, and the hanger 14 m from the left tower has the same length
Common mistakes
- Solving y = 0 to find the towers. The cable never reaches the deck: 18x2 − 2x + 10 = 0 has no real roots. The towers are where the cable is at the height of the tower tops, y = 10.
- Taking the other hanger of the same length to be at x = 4, twice as far from the tower. Equal heights are at equal distances from the axis of symmetry x = 8, so the matching hanger is at 8 + 6 = 14.
More quadratic graphs and coordinate geometry problems, worked step by step →