Sketching quadratic graphs — key features — Mathematics

Before this

This leads to

Sketching cubics and higher polynomials
Stationary points — finding and classifying · coming soon
Range and maximum height of a projectile

Two tables, and neither one finds the bottom

Priya and Tomás both need the graph of $y = x^{2} - 4 x - 5$ . Priya makes a table for $x = 3, 4, 5, 6, 7$ and gets $y = - 8, - 5, 0, 7, 16$ . Tomás uses $x = - 3, - 2, - 1, 0, 1$ and gets $y = 16, 7, 0, - 5, - 8$ .

Tomás's points (hollow) only fall. Priya's points (filled) only rise. The dashed curve is the true graph. Its lowest point lies between the two tables.

Both tables are correct, and both miss the most important point. This is the disruption. A table only answers the questions you thought to ask. If you choose the wrong values of $x$ , you miss the turn. Sketching works the other way round. You read the key features from the equation first. Then you know where to look.

A sketch is not a careful plot on graph paper. It is a drawing of the right shape, in the right place, with the key points labelled with their exact coordinates.

Every quadratic graph has the same shape

The graph of $y = a x^{2} + b x + c$ , with $a \neq = 0$ , is called a parabola. The lecture on drawing straight lines found that $y = x^{2}$ has values $0, 1, 4, 9, 16$ at $x = 0, 1, 2, 3, 4$ , with steps $1, 3, 5, 7$ , so its graph is a curve. It promised that curves like this would come later. Here they are.

Look at those steps again. Each step is exactly 2 more than the one before. So the graph gets steeper at a steady rate, and that steady bending gives a parabola its smooth, even shape.

Every quadratic behaves like this. Take $y = x^{2} - 4 x - 5$ at $x = 0, 1, 2, 3, 4, 5$ . The values are $- 5, - 8, - 9, - 8, - 5, 0$ . The steps are $- 3, - 1, 1, 3, 5$ . Again, each step is 2 more than the last. The curve falls, falls less, turns, and rises.

In general the steps change by $2 a$ each time, so the sign of $a$ decides which way the curve bends. There is a second way to see this. Far from the origin, the $x^{2}$ term is much bigger than the others. At $x = 100$ , the terms of $x^{2} - 4 x - 5$ are $10000$ , $- 400$ and $- 5$ . The same happens at $x = - 100$ , because $(- 100)^{2}$ is also positive. So far out on both sides, $y$ has the same sign as $a$ .

If $a > 0$ , both arms go up. The graph is a U-shape, and its turning point is a minimum.
If $a < 0$ , both arms go down. The graph is an ∩-shape, and its turning point is a maximum.

The completing-the-square lecture showed that any quadratic can be written as $a (x - h)^{2} + k$ . That is the graph of $y = x^{2}$ , stretched by $a$ and moved so that its turning point is at $(h, k)$ . There is only one shape. Only its width, direction and position change.

Where the curve meets the $y$ -axis

Every point on the $y$ -axis has $x = 0$ . Put $x = 0$ into $y = a x^{2} + b x + c$ , and the first two terms vanish. So the curve always crosses the $y$ -axis at $(0, c)$ .

Stop and think

Earlier lectures found the mirror line halfway between the two roots. Now take $y = x^{2} + 2 x + 5$ . Its discriminant is $2^{2} - 4 \times 1 \times 5 = - 16$ , so it has no roots. Does it still have a mirror line? How could you find it?

Show the answer

Yes. Every parabola is symmetric. The roots are just one pair of points at the same height, and any other pair works equally well. You need a pair that every quadratic has, roots or not. The next section finds one.

The mirror line, found without the roots

We know one point: $(0, c)$ . Where else does the curve reach the height $c$ ? Set $y = c$ and solve:

a x^{2} + b x + c a x^{2} + b x x (a x + b) = c = 0 = 0

So $x = 0$ or $x = - \frac{b}{a}$ . The curve passes through a second point at height $c$ : $(- \frac{b}{a}, c)$ .

Two different points on a parabola at the same height are mirror images. This is because $a (x - h)^{2} + k$ gives the same value when $x$ is the same distance from $h$ on either side, and only then. So the mirror line is halfway between them, at the average of $0$ and $- \frac{b}{a}$ :

x = \frac{0 + ( - \frac{b}{a} )}{2} = - \frac{b}{2 a}

That is it. That is where $x = - \frac{b}{2 a}$ comes from. It is not a formula to learn by heart. It is the average of two places at the same height. This line is the axis of symmetry, and the argument works for every quadratic, with roots or without.

y = x^{2} + 2 x + 5

has no roots, but it reaches height 5 twice: at

x = 0

and at

x = - \frac{b}{a} = - 2

. The mirror line is halfway between them, at

x = - 1

The second point, $(- \frac{b}{a}, c)$ , is also a free point for your sketch. Call it the mirror point of the $y$ -intercept.

The vertex: the turning point

The turning point is called the vertex. It lies on the axis of symmetry, so its $x$ -coordinate is $- \frac{b}{2 a}$ . To find its $y$ -coordinate, substitute that value into the equation. For $y = x^{2} + 2 x + 5$ , the axis is $x = - 1$ , and

y = (- 1)^{2} + 2 (- 1) + 5 = 1 - 2 + 5 = 4

So the vertex is $(- 1, 4)$ , a minimum, because $a = 1$ is positive. The discriminant lecture gave the same height another way: $- \frac{b ^{2} - 4 a c}{4 a} = - \frac{- 16}{4} = 4$ ✓. Substitution is usually quicker.

Where the curve meets the $x$ -axis

Every point on the $x$ -axis has $y = 0$ , so the roots come from solving $a x^{2} + b x + c = 0$ . Factorise when you can see the brackets. Use the formula or completing the square when you cannot. The discriminant, $b^{2} - 4 a c$ , tells you first what to expect: positive, two roots; zero, one repeated root, with the vertex on the axis; negative, no real roots. Two roots are mirror images, so they sit the same distance either side of the axis. That is a free check on every sketch.

Sketching

y = a x^{2} + b x + c

Shape: $a > 0$ gives a U-shape with a minimum; $a < 0$ gives an ∩-shape with a maximum.

$y$ -intercept: $(0, c)$ , and its mirror point $(- \frac{b}{a}, c)$ .

Axis of symmetry: $x = - \frac{b}{2 a}$ , halfway between any two points at the same height.

Vertex: substitute $x = - \frac{b}{2 a}$ to find its $y$ -coordinate.

Roots: solve $a x^{2} + b x + c = 0$ when $b^{2} - 4 a c \geq 0$ .

Then draw one smooth, symmetric curve through the marked points, and label every point with its coordinates.

What the middle number does

The constant $c$ says where the curve crosses the $y$ -axis. The number $a$ says which way it opens, and how narrow it is. The middle number, $b$ , is the puzzle. It moves the vertex, but in which direction? Make a prediction before you read on.

Predict — which way does b move the vertex?

Both curves are U-shapes through (0, 3). One has its vertex to the left of the y-axis, the other to the right. Choose the statement you believe, then read the explanation.

The two curves y = x² + 4x + 3 and y = x² − 4x + 3 both pass through (0, 3). The one with +4x has its vertex on the left, at (−2, −1). The one with −4x has its vertex on the right, at (2, −1).

The curve with

+ 4 x

is rising as it crosses the

y

-axis, so its lowest point is already behind it, on the left. The curve with

- 4 x

is still falling, so its lowest point is ahead, on the right.

Here is the reason. The number $b$ is the gradient of the curve where it crosses the $y$ -axis. The gradient idea from the straight-line lectures proves it. Take two points on the curve, the same distance $h$ either side of the $y$ -axis:

y (h) y (- h) = a h^{2} + bh + c = a h^{2} - bh + c

The difference in height is $y (h) - y (- h) = 2 bh$ . The difference in $x$ is $2 h$ . So the line joining the two points has gradient $\frac{2 bh}{2 h} = b$ . The answer does not depend on $h$ . However close the two points are, the line joining them has gradient $b$ . So at the $y$ -axis, the curve itself is heading with gradient $b$ .

In $y = x^{2} + 4 x + 3$ , the curve is rising with gradient 4 at the $y$ -axis. A U-shape that is already rising has passed its lowest point. So the vertex is on the left. In $y = x^{2} - 4 x + 3$ , the curve is still falling at the $y$ -axis, so its lowest point is still ahead, on the right. So a positive $b$ moves the vertex of a U-shape to the left, which is what the minus sign in $- \frac{b}{2 a}$ says.

Try all three numbers together in the graph builder below.

Graph builder — what a, b and c each do

The curve starts as y = x² − 4x − 5, from Worked Example 1. Change one number at a time. First move only c and watch the dashed axis. Then move only b and watch the point (0, c). Last, move a down through 0 and watch the curve turn over.

a =

b =

c =

What each number does, starting from y = x² − 4x − 5 (vertex (2, −9)):

Change	What moves	What stays fixed
c from −5 to 3: y = x² − 4x + 3	The whole curve rises by 8. Vertex (2, −1).	The axis x = 2.
b from −4 to 4: y = x² + 4x − 5	The vertex moves to (−2, −9), on the other side.	The point (0, −5), now rising with gradient 4.
b from −4 to 0: y = x² − 5	The vertex moves onto the y-axis, at (0, −5).	The point (0, −5), now flat.
a from 1 to 2: y = 2x² − 4x − 5	Narrower curve. Axis x = 1, vertex (1, −7).	The point (0, −5) and its gradient −4.
a from 1 to −1: y = −x² − 4x − 5	An ∩-shape. Axis x = −2, maximum (−2, −1), no real roots.	The point (0, −5) and its gradient −4.
a = 0: y = −4x − 5	No longer a parabola: a straight line.	Gradient −4 through (0, −5).

What should you have seen? Changing $c$ lifts or lowers the whole curve, and the axis never moves, because $- \frac{b}{2 a}$ contains no $c$ . Changing $b$ keeps $(0, c)$ fixed and swings the curve around it. Changing $a$ makes the curve narrower or wider. At $a = 0$ it becomes the straight line $y = b x + c$ , with gradient $b$ through $(0, c)$ : the direction of every one of these parabolas at the $y$ -axis.

Worked examples

Worked Example 1 — Back to Priya and Tomás

Sketch $y = x^{2} - 4 x - 5$ . Label all the key features.

Shape. $a = 1$ is positive, so the graph is a U-shape with a minimum.

$y$ -intercept. $c = - 5$ , so the curve passes through $(0, - 5)$ . Its mirror point is at $x = - \frac{b}{a} = - \frac{- 4}{1} = 4$ , so $(4, - 5)$ is also on the curve.

Axis. $x = - \frac{b}{2 a} = - \frac{- 4}{2 \times 1} = 2$ . This is halfway between 0 and 4 ✓.

Vertex. $y = 2^{2} - 4 (2) - 5 = 4 - 8 - 5 = - 9$ . The minimum is $(2, - 9)$ .

Roots. $x^{2} - 4 x - 5 = (x - 5) (x + 1) = 0$ , so $x = 5$ or $x = - 1$ . Check: halfway between them is $\frac{5 + ( - 1 )}{2} = 2$ , the axis ✓.

Gradient check. $b = - 4$ , so the curve is falling as it crosses the $y$ -axis. The minimum should be to the right of the $y$ -axis, and it is ✓.

The finished sketch: five labelled points and the mirror line. The turn that Priya and Tomás missed is found first.

Worked Example 2 — An upside-down parabola

Sketch $y = - x^{2} + 2 x + 8$ .

Shape. $a = - 1$ is negative, so the graph is an ∩-shape with a maximum.

$y$ -intercept. $(0, 8)$ . Mirror point: $x = - \frac{b}{a} = - \frac{2}{- 1} = 2$ , so $(2, 8)$ .

Axis. $x = - \frac{2}{2 \times ( - 1 )} = - \frac{2}{- 2} = 1$ . Halfway between 0 and 2 ✓.

Vertex. $y = - (1)^{2} + 2 (1) + 8 = - 1 + 2 + 8 = 9$ . The maximum is $(1, 9)$ .

Roots. Solve $- x^{2} + 2 x + 8 = 0$ . Multiply every term by $- 1$ . This does not change the solutions, because $0 \times (- 1)$ is still $0$ . You get $x^{2} - 2 x - 8 = 0$ , so $(x - 4) (x + 2) = 0$ , and $x = 4$ or $x = - 2$ . Halfway between them is 1 ✓.

Gradient check. $b = 2$ : rising at the $y$ -axis. An ∩-shape that is still rising has not reached its top, so the maximum is to the right ✓. The sketch is an ∩-shape through $(- 2, 0)$ , $(0, 8)$ , $(1, 9)$ , $(2, 8)$ and $(4, 0)$ .

Worked Example 3 — No roots at all

Sketch $y = 2 x^{2} + 4 x + 5$ .

Shape. $a = 2$ , positive: a U-shape, and narrower than $y = x^{2}$ , because the steps grow twice as fast.

$y$ -intercept. $(0, 5)$ . Mirror point: $x = - \frac{4}{2} = - 2$ , so $(- 2, 5)$ .

Axis and vertex. $x = - \frac{4}{2 \times 2} = - 1$ . Then $y = 2 (- 1)^{2} + 4 (- 1) + 5 = 2 - 4 + 5 = 3$ . The minimum is $(- 1, 3)$ .

Roots. $b^{2} - 4 a c = 16 - 40 = - 24$ , which is negative. There are no real roots.

Does that agree with the picture? Yes. The lowest point of the curve is at height 3, above the $x$ -axis, so the curve can never come down to the axis ✓.

The sketch: a U-shape entirely above the $x$ -axis, through $(- 2, 5)$ , $(- 1, 3)$ and $(0, 5)$ , labelled "no real roots". The mirror point gives this sketch three points instead of two.

Worked Example 4 — Roots that are not whole numbers

Sketch $y = x^{2} - 6 x + 4$ .

Shape, intercept, mirror point. U-shape. $(0, 4)$ , and its mirror point at $x = - \frac{- 6}{1} = 6$ , so $(6, 4)$ .

Axis and vertex. $x = 3$ . Then $y = 9 - 18 + 4 = - 5$ . The minimum is $(3, - 5)$ .

Roots. $b^{2} - 4 a c = 36 - 16 = 20$ , positive, so there are two roots. No pair of whole numbers multiplies to 4 and adds to $- 6$ , so the quadratic does not factorise. The vertex already gives the completed square: $y = (x - 3)^{2} - 5$ . Setting this to zero gives $(x - 3)^{2} = 5$ , so

x = 3 \pm 5

Label the roots exactly, as $(3 - 5, 0)$ and $(3 + 5, 0)$ . For placing them on the sketch, $3 - 5 \approx 0.76$ and $3 + 5 \approx 5.24$ (3 significant figures, written 3 s.f.).

Check. The roots are $5$ either side of the axis ✓. The curve is at height 4 on the $y$ -axis and falling ( $b = - 6$ ), so it should cross the $x$ -axis soon after, and it does, near 0.76 ✓.

Three traps

Trap 1 — Losing the minus sign in

- \frac{b}{2 a}

For $y = x^{2} - 4 x - 5$ , the number $b$ is $- 4$ , not 4. So $- b = 4$ , and the axis is $x = 2$ . Writing $x = \frac{- 4}{2} = - 2$ puts the vertex on the wrong side. The gradient check catches this at once: $b = - 4$ means the curve is falling at the $y$ -axis, so the minimum must be to the right.

Trap 2 — Squaring a negative number wrongly

To find $y$ on $y = - x^{2} + 2 x + 8$ at $x = - 3$ , you need $- (- 3)^{2}$ . The square comes first: $(- 3)^{2} = 9$ , and then the minus sign gives $- 9$ . So $y = - 9 - 6 + 8 = - 7$ . Writing $+ 9$ here is a very common slip. On a calculator, type the brackets: $(- 3)^{2}$ , not $- 3^{2}$ .

Trap 3 — A pointed tip, or arms that stop

A parabola is round at the vertex, never pointed like a V. Its arms keep going: they do not stop at the roots, and they never curve back in. Draw the two sides as mirror images about the dashed axis.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Sketch $y = x^{2} + 2 x - 8$ . Label the $y$ -intercept, its mirror point, the vertex and the roots.

Hint 1
Start with the shape, then read $c$ . The mirror point is at $x = - \frac{b}{a}$ .

Hint 2
The axis is $x = - \frac{2}{2} = - 1$ . For the roots, find two numbers that multiply to $- 8$ and add to 2.

Full answer
$a = 1$ : U-shape, minimum. $y$ -intercept $(0, - 8)$ ; mirror point at $x = - 2$ , so $(- 2, - 8)$ . Axis $x = - 1$ . Vertex: $y = 1 - 2 - 8 = - 9$ , so $(- 1, - 9)$ . Roots: $(x + 4) (x - 2) = 0$ , so $(- 4, 0)$ and $(2, 0)$ . Check: halfway between $- 4$ and 2 is $- 1$ ✓. Gradient check: $b = 2$ , rising at the $y$ -axis, so the minimum is on the left ✓.
Sketch $y = - 2 x^{2} + 8 x - 3$ . Give the roots exactly and to 3 s.f.

Hint 1
$a = - 2$ , so the curve is an ∩-shape. The axis is $x = - \frac{8}{2 \times ( - 2 )}$ .

Hint 2
The discriminant is $64 - 24 = 40$ . Use the quadratic formula, and simplify $40 = 210$ .

Full answer
∩-shape, maximum. $y$ -intercept $(0, - 3)$ ; mirror point at $x = - \frac{8}{- 2} = 4$ , so $(4, - 3)$ . Axis $x = 2$ . Vertex: $y = - 2 (4) + 16 - 3 = 5$ , so the maximum is $(2, 5)$ . Roots: $x = \frac{- 8 \pm 40}{- 4} = 2 \mp \frac{10}{2}$ , so $x = 2 - \frac{10}{2} \approx 0.419$ and $x = 2 + \frac{10}{2} \approx 3.58$ . Check: both are $\frac{10}{2}$ from the axis $x = 2$ ✓. The curve starts below the axis at $(0, - 3)$ and is rising ( $b = 8$ ), so it crosses the axis soon after, near 0.42 ✓.
A parabola crosses the $x$ -axis at $- 1$ and $5$ , and the $y$ -axis at $(0, - 10)$ . Find its equation and its vertex.

Hint 1
Roots $- 1$ and 5 mean the quadratic has factors $(x + 1)$ and $(x - 5)$ . There may also be a number $a$ in front.

Hint 2
Write $y = a (x + 1) (x - 5)$ . Put in $x = 0$ , $y = - 10$ to find $a$ .

Full answer
$y = a (x + 1) (x - 5)$ . At $x = 0$ : $- 10 = a (1) (- 5) = - 5 a$ , so $a = 2$ . Expanding, $y = 2 (x^{2} - 4 x - 5) = 2 x^{2} - 8 x - 10$ . The axis is halfway between the roots, at $x = 2$ . The vertex: $y = 2 (4) - 16 - 10 = - 18$ , so $(2, - 18)$ . Check with the formula: $- \frac{b}{2 a} = - \frac{- 8}{4} = 2$ ✓. This curve is Worked Example 1 stretched by the factor 2: the roots stay put, and every height doubles, so $- 9$ becomes $- 18$ .
A farmer has 24 m of fencing. She builds a rectangular pen against a long wall, so she fences only three sides: two sides of width $x$ metres, and one side parallel to the wall. Write the area $A$ in terms of $x$ . Sketch $A$ against $x$ , and use the sketch to find the largest possible area. Which part of your sketch makes sense for the pen? (An invented example.)

Hint 1
Two widths use $2 x$ metres, so the side parallel to the wall is $24 - 2 x$ . The area is width × length.

Hint 2
$A = x (24 - 2 x) = - 2 x^{2} + 24 x$ . Its roots are easy to read from the factorised form.

Full answer
$A = x (24 - 2 x) = - 2 x^{2} + 24 x$ . This is an ∩-shape ( $a = - 2$ ), so it has a maximum. The roots are $x = 0$ and $x = 12$ , where the pen has no area. The axis is halfway, at $x = 6$ (and $- \frac{24}{2 \times ( - 2 )} = 6$ ✓). The maximum is $A = 6 \times (24 - 12) = 72$ m², with a pen 6 m wide and 12 m long. Check two neighbours: $x = 5$ gives $5 \times 14 = 70$ , and $x = 7$ gives $7 \times 10 = 70$ , both smaller and equal, as the symmetry says ✓. Only the arch between $x = 0$ and $x = 12$ makes sense: outside it, a length or the area would be negative. The vertex answers two questions at once: the best area (72 m²) and how to get it ( $x = 6$ ).

The bigger idea

The vertex is an optimum: the smallest or largest value the quadratic can take. The fencing question in Try it 4 is one example. A ball thrown into the air is another: if air resistance is ignored, its path is a parabola, and the vertex is the top of its flight.

The two-point trick works at any point, not only at $x = 0$ . Join the points at $x - h$ and $x + h$ , and the gradient comes out as $2 a x + b$ , again with no $h$ left. At the vertex the curve is flat, so $2 a x + b = 0$ , which gives $x = - \frac{b}{2 a}$ a third time. Finding where a curve is flat is the central move of differentiation, the first great idea of calculus. The next lecture uses the same features to sketch cubics, which can turn twice.

Key concepts in this lesson

The graph of $y = a x^{2} + b x + c$ is a parabola: a U-shape with a minimum when $a > 0$ , and an ∩-shape with a maximum when $a < 0$ .
The curve crosses the $y$ -axis at $(0, c)$ and reaches the same height again at the mirror point $(- \frac{b}{a}, c)$ .
The axis of symmetry $x = - \frac{b}{2 a}$ is halfway between those two points, so it can be found even when the curve has no roots.
The vertex lies on the axis, and its $y$ -coordinate is found by substituting $x = - \frac{b}{2 a}$ into the equation.
The number $b$ is the gradient of the curve where it crosses the $y$ -axis, which shows which side of the $y$ -axis the vertex lies on.