Introduction to Quadratic Relations

In Grade 9 you studied linear relations, whose graphs are straight lines. A quadratic relation is the next step up: its equation has an $x^2$ term, and its graph is a U-shaped curve called a parabola. Parabolas are everywhere: the path of a thrown ball, the shape of a satellite dish, the arch of a bridge. This page shows you how to recognize a quadratic relation and how to describe its graph.

Key ideas

What is a quadratic relation?

A quadratic relation has an equation that can be written in the form

$$y = ax^2 + bx + c, \qquad a \ne 0$$

where $a$, $b$ and $c$ are numbers. This is called standard form. The highest power of $x$ is $2$. The condition $a \ne 0$ matters: if $a = 0$, the $x^2$ term disappears and you’re left with a linear relation.

• $y = 3x^2 - 2x + 1$ is quadratic ($a = 3$, $b = -2$, $c = 1$).
• $y = 5 - x^2$ is quadratic ($a = -1$, $b = 0$, $c = 5$).
• $y = 2x(x - 3)$ is quadratic too: expand it to get $y = 2x^2 - 6x$.
• $y = 4x - 1$ is linear, and $y = x^3$ is neither.

First and second differences

You can also spot a quadratic relation in a table of values, as long as the $x$-values go up by the same step each time.

• The first differences are the changes in $y$ from one row to the next.
• The second differences are the changes in the first differences.

For a linear relation, the first differences are constant. For a quadratic relation, the first differences change, but the second differences are constant (and not zero).

Here is the table for $y = x^2 - 2x - 3$:

$x$	$y$	First differences	Second differences
$-2$	$5$
$-1$	$0$	$0 - 5 = -5$
$0$	$-3$	$-3$	$-3 - (-5) = 2$
$1$	$-4$	$-1$	$2$
$2$	$-3$	$1$	$2$
$3$	$0$	$3$	$2$
$4$	$5$	$5$	$2$

The second differences are all $2$, so the relation is quadratic. (A bonus pattern: when the $x$-values go up by $1$, the second difference is always $2a$. Here $a = 1$, so it’s $2$.)

The parabola and its key features

The graph of every quadratic relation is a parabola. These are the features you’ll describe again and again:

Feature	What it is
direction of opening	up if $a \gt 0$, down if $a \lt 0$
vertex	the turning point: the lowest point if it opens up, the highest if it opens down
axis of symmetry	the vertical line through the vertex; write it as an equation, like $x = 1$
minimum or maximum value	the $y$-coordinate of the vertex
$y$-intercept	where the graph crosses the $y$-axis; set $x = 0$, which gives $y = c$
zeros ($x$-intercepts)	where the graph crosses the $x$-axis ($y = 0$); there can be $2$, $1$ or $0$ of them

The key features of $y = x^2 - 2x - 3$. The vertex $(1, -4)$ gives a minimum value of $-4$.

Symmetry

A parabola is the same on both sides of its axis of symmetry. That gives you useful shortcuts:

• Two points with the same $y$-value are the same distance from the axis, so the axis is halfway between them.
• In particular, the axis of symmetry is halfway between the two zeros. In the graph above, the zeros are $-1$ and $3$, and $\dfrac{-1 + 3}{2} = 1$.
• In a table of values, the $y$-values repeat in mirror order around the vertex: $5, 0, -3, -4, -3, 0, 5$.

Data and the curve of best fit

Real measurements can also follow a quadratic pattern. For example, you could toss a ball straight up under a motion sensor, roll a can down a ramp and time it, or look up data from a source like Statistics Canada. To check whether a quadratic model fits:

1. Make a scatter plot of the data.
2. If the points follow a U shape (or an upside-down U), and the second differences are roughly constant, a quadratic model is reasonable. Real data never fits perfectly.
3. Draw a curve of best fit: one smooth parabola that passes through or close to as many points as possible, with the points balanced on either side. Don’t join the dots with straight segments.
4. Use the curve to estimate values, such as the vertex or a zero.

Here is a sample data set from a ball tossed upward from a height of about $1.2$ m:

Time (s)	$0$	$0.2$	$0.4$	$0.6$	$0.8$	$1.0$	$1.2$	$1.4$	$1.6$
Height (m)	$1.2$	$2.6$	$3.5$	$4.1$	$4.3$	$4.1$	$3.5$	$2.5$	$1.1$

The data points and a curve of best fit. The curve, not the dots, is used to estimate when the ball reaches the floor.

Worked examples

Example 1: Linear, quadratic, or neither?

Use differences to decide whether each table shows a linear relation, a quadratic relation, or neither.

$x$	$0$	$1$	$2$	$3$	$4$
Table A: $y$	$3$	$5$	$7$	$9$	$11$
Table B: $y$	$1$	$2$	$5$	$10$	$17$
Table C: $y$	$1$	$3$	$9$	$27$	$81$

Solution. The $x$-values go up by $1$ each time, so differences work.

Table A. First differences: $2, 2, 2, 2$. They’re constant, so the relation is linear.

Table B. First differences: $1, 3, 5, 7$. They change, so find the second differences: $2, 2, 2$. Constant, so the relation is quadratic.

Table C. First differences: $2, 6, 18, 54$. Second differences: $4, 12, 36$. Neither set is constant, so the relation is neither linear nor quadratic. (Each $y$-value is $3$ times the one before, so this relation is exponential.)

Example 2: From a table of values to the key features

Make a table of values for $y = -x^2 + 4x + 5$ using $x = -1$ to $5$. Graph it, and state the key features.

Solution. Substitute each $x$-value. Use brackets around negatives, and remember that $-x^2$ means “square $x$, then make it negative”. For example, at $x = -1$:

$$y = -(-1)^2 + 4(-1) + 5 = -1 - 4 + 5 = 0$$

$x$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$0$	$5$	$8$	$9$	$8$	$5$	$0$

Plot the points and join them with a smooth curve. Reading the table:

• Direction of opening: down, since $a = -1 \lt 0$.
• Vertex: $(2, 9)$. The $y$-values rise to $9$ and then repeat in mirror order.
• Axis of symmetry: $x = 2$.
• Maximum value: $9$.
• $y$-intercept: $5$ (the value at $x = 0$, which is $c$).
• Zeros: $-1$ and $5$ (where $y = 0$).

Check the symmetry: the zeros $-1$ and $5$ average to $\dfrac{-1 + 5}{2} = 2$, which is the axis. ✓

Example 3: Using symmetry

A parabola has zeros at $-3$ and $5$ and a minimum value of $-8$. It passes through the point $(-1, -6)$.

• (a) Find the equation of the axis of symmetry and the coordinates of the vertex.
• (b) Find another point on the parabola with a $y$-coordinate of $-6$.

Solution.

(a) The axis is halfway between the zeros:

$$x = \frac{-3 + 5}{2} = 1$$

The axis of symmetry is $x = 1$. The vertex is on the axis, and its $y$-coordinate is the minimum value, so the vertex is $(1, -8)$.

(b) The point $(-1, -6)$ is $2$ units to the left of the axis $x = 1$. Its mirror image is $2$ units to the right, at $x = 3$, with the same $y$-value. So $(3, -6)$ is on the parabola.

Example 4: A curve of best fit

Use the tossed-ball data and the curve of best fit above.

• (a) Show that a quadratic model is reasonable.
• (b) Estimate the maximum height and when it happens.
• (c) Estimate when the ball reaches the floor.

Solution.

(a) The times go up by $0.2$ s each time. First differences of the heights:

$$1.4,\ 0.9,\ 0.6,\ 0.2,\ -0.2,\ -0.6,\ -1.0,\ -1.4$$

Second differences:

$$-0.5,\ -0.3,\ -0.4,\ -0.4,\ -0.4,\ -0.4,\ -0.4$$

They’re nearly constant (about $-0.4$), and the scatter plot is an upside-down U. A quadratic model is reasonable, and it opens down (negative second differences).

(b) The highest point of the curve is near $(0.8, 4.3)$. The maximum height is about $4.3$ m, about $0.8$ s after the toss.

(c) Extend the curve down to the time axis. It crosses at about $1.7$ s, so the ball reaches the floor after about $1.7$ s. This is an estimate, read from a curve drawn through measured data.

Common mistakes

Taking differences when the $x$-steps aren’t equal. Differences only tell you the type of relation if the $x$-values go up by the same amount each time. Check the $x$-column first.

Stopping at the first differences. First differences that change don’t mean “not quadratic”. Go one step further and check the second differences.

Writing the axis of symmetry as a number. The axis is a line, so write it as an equation: $x = 2$, not just $2$.

Mixing up the vertex and the maximum value. The vertex is a point, $(2, 9)$. The maximum value is just its $y$-coordinate, $9$.

Evaluating $-x^2$ incorrectly. For $x = -3$, $-x^2 = -(-3)^2 = -9$, not $+9$. Square first, then apply the negative sign.

Joining data points with straight segments. A curve of best fit is one smooth parabola through the middle of the data. It doesn’t have to pass through every point.

Practice

1. (Warm-up) Which of these relations are quadratic?

• (a) $y = 5 - 2x^2$
• (b) $y = 3x + 7$
• (c) $y = x(x + 4)$
• (d) $y = 2^x$

Solution

(a) Quadratic: $y = -2x^2 + 5$, with $a = -2$.

(b) Linear: the highest power of $x$ is $1$.

(c) Quadratic: expanding gives $y = x^2 + 4x$.

(d) Neither: $x$ is in the exponent, so this is exponential.

2. (Warm-up) Is this relation linear, quadratic, or neither? Explain.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$y$	$2$	$3$	$6$	$11$	$18$	$27$

Solution

First differences: $1, 3, 5, 7, 9$. Second differences: $2, 2, 2, 2$.

The second differences are constant (and not zero), so the relation is quadratic.

3. (Warm-up) A parabola opens down and has its vertex at $(-2, 7)$. State the equation of its axis of symmetry and its maximum or minimum value.

Solution

The axis of symmetry is the vertical line through the vertex: $x = -2$.

It opens down, so the vertex is the highest point. The maximum value is $7$.

4. (Core) Make a table of values for $y = 2x^2 - 8x + 3$ using $x = 0$ to $4$. Check that the second differences are constant, then state the vertex, the axis of symmetry, the minimum or maximum value, and the $y$-intercept.

Solution

$x$	$0$	$1$	$2$	$3$	$4$
$y$	$3$	$-3$	$-5$	$-3$	$3$

For example, at $x = 1$: $y = 2(1)^2 - 8(1) + 3 = 2 - 8 + 3 = -3$.

First differences: $-6, -2, 2, 6$. Second differences: $4, 4, 4$. Constant, as expected (and $4 = 2a$).

The $y$-values repeat in mirror order around $x = 2$, so:

• vertex $(2, -5)$
• axis of symmetry $x = 2$
• minimum value $-5$ (it opens up, since $a = 2 \gt 0$)
• $y$-intercept $3$

5. (Core) A parabola passes through $(-4, 10)$ and $(2, 10)$, and its vertex has a $y$-coordinate of $-8$. Find the axis of symmetry and the vertex. Does the parabola have a maximum or a minimum?

Solution

The two points have the same $y$-value, so the axis is halfway between them:

$$x = \frac{-4 + 2}{2} = -1$$

The axis is $x = -1$ and the vertex is $(-1, -8)$.

The vertex is below the points $(-4, 10)$ and $(2, 10)$, so the parabola rises from the vertex: it opens up and has a minimum value of $-8$.

6. (Core) A parabola has zeros at $-6$ and $2$ and a $y$-intercept of $6$.

• (a) Find the equation of the axis of symmetry.
• (b) Does it open up or down? Explain.
• (c) Use symmetry to find another point on the parabola with a $y$-coordinate of $6$.

Solution

(a) $x = \dfrac{-6 + 2}{2} = -2$, so the axis is $x = -2$.

(b) The point $(0, 6)$ lies between the zeros and above the $x$-axis. So between its zeros the parabola is above the $x$-axis, which means it rises from one zero and comes back down to the other. It opens down.

(c) $(0, 6)$ is $2$ units right of the axis $x = -2$. Its mirror image is $2$ units left, at $x = -4$: the point $(-4, 6)$.

7. (Core) The table gives the height $y$ (in metres) of a footbridge arch at a horizontal distance $x$ (in metres) from one end.

$x$ (m)	$0$	$2$	$4$	$6$	$8$
$y$ (m)	$0$	$12$	$16$	$12$	$0$

• (a) Is the arch modelled by a quadratic relation? Explain.
• (b) State the zeros, the vertex, and the maximum height of the arch.

Solution

(a) The $x$-values go up by $2$ each time, so differences work. First differences: $12, 4, -4, -12$. Second differences: $-8, -8, -8$. They’re constant, so yes, it’s quadratic (and it opens down).

(b) The zeros are $0$ and $8$ (where $y = 0$): the arch meets the ground at each end. The vertex is $(4, 16)$, halfway between the zeros, so the maximum height is $16$ m.

8. (Challenge) A driving instructor records the braking distance of a car at different speeds on dry pavement.

Speed (km/h)	$20$	$40$	$60$	$80$	$100$
Braking distance (m)	$2.5$	$9.8$	$22.4$	$39.5$	$62.1$

• (a) Show that a quadratic model is reasonable.
• (b) Use the pattern in the differences to estimate the braking distance at $120$ km/h.

Solution

(a) The speeds go up by $20$ km/h each time. First differences: $7.3, 12.6, 17.1, 22.6$. Second differences: $5.3, 4.5, 5.5$. These are roughly constant (about $5$), so a quadratic model is reasonable.

(b) If the second difference stays at about $5$, the next first difference is about $22.6 + 5 = 27.6$. So the braking distance at $120$ km/h is about

$$62.1 + 27.6 = 89.7$$

or roughly $90$ m. (Using a curve of best fit gives a similar estimate, around $89$ to $90$ m. Answers close to this are fine, since the data isn’t perfect.) Notice that going from $100$ km/h to $120$ km/h adds almost $28$ m of braking distance.

9. (Challenge) A table of values for a quadratic relation starts with these rows. The $x$-values go up by $1$.

$x$	$0$	$1$	$2$
$y$	$4$	$1$	$0$

Use the fact that the second differences are constant to find $y$ when $x = 3$ and $x = 4$. Then state the vertex.

Solution

First differences: $1 - 4 = -3$ and $0 - 1 = -1$. Second difference: $-1 - (-3) = 2$.

Keep the second difference at $2$. The next first differences are $-1 + 2 = 1$ and $1 + 2 = 3$. So:

$$y(3) = 0 + 1 = 1, \qquad y(4) = 1 + 3 = 4$$

The $y$-values are $4, 1, 0, 1, 4$, which repeat in mirror order around $x = 2$. The vertex is $(2, 0)$. (The relation is $y = (x - 2)^2$.)