Parabola through three points

What is it about?

Two points determine a line. A parabola $y = ax^2 + bx + c$ needs three points, because three numbers $a$ , $b$ , $c$ are unknown. Substituting each point $P_i(x_i \mid y_i)$ gives one equation, so you get a system of linear equations in $a$ , $b$ , $c$ :

x_i^2\,a + x_i\,b + c = y_i \qquad (i = 1, 2, 3)

It is solved by elimination. First eliminate $c$ ( $\text{II} - \text{I}$ , $\text{III} - \text{I}$ ). Then eliminate $b$ , which gives $a$ . Finally back-substitute to get $b$ and $c$ . If one point lies on the $y$ -axis, $c$ can be read off at once.

Shortcuts: if one point is the vertex, use $y = a(x - x_S)^2 + y_S$ . If two points are zeros, use $y = a(x - x_1)(x - x_2)$ .

Special cases: two points with the same $x$ but different $y$ give a contradiction, so no function passes through them. If two points coincide, infinitely many parabolas pass through them. Three points on a line give $a = 0$ : a line, not a parabola.

Try it

Drag the points: the parabola always passes through all three. Load example
Click Step-by-step solution and see where the coloured numbers come from.
Load the points $(0 \mid 0)$ , $(1 \mid 1)$ and $(2 \mid 4)$ . Which parabola is it? Why does equation I only say “c = 0”? Load example
Points on a line: drag $P_2$ slightly up and back again. What happens to $a$ ? Load example
Same $x$ -value: why is there no solution? Which equation does II − I give? Load example
Put two points on top of each other. How many parabolas are there now? Load example
Choose the method Vertex form: here $P_2$ is the vertex. Compare the length of the working with the system of equations. Load example
Choose the application Path of a thrown ball and throw the ball. Where does it land? Load example

Tasks

Task 1

Load task

Find the parabola through $A(0 \mid 1)$ , $B(1 \mid 0)$ , $C(2 \mid 3)$ .

Show solution

$c = 1$ . Then $a + b = -1$ and $4a + 2b = 2$ , so $a = 2$ and $b = -3$ : $f(x) = 2x^2 - 3x + 1$ . Show

Task 2

A parabola has its vertex at $S(2 \mid -3)$ and passes through $P(4 \mid 1)$ . Find its equation in vertex form and in general form.

Show solution

$y = a(x - 2)^2 - 3$ . Substituting $P$ gives $1 = a \cdot 2^2 - 3$ , so $a = 1$ : $f(x) = (x - 2)^2 - 3 = x^2 - 4x + 1$ . Show

Task 3

A parabola has the zeros $-2$ and $4$ and passes through $(2 \mid 4)$ . Find its equation.

Show solution

$y = a(x + 2)(x - 4)$ . Substituting gives $4 = a \cdot 4 \cdot (-2) = -8a$ , so $a = -\frac{1}{2}$ : $f(x) = -\frac{1}{2}x^2 + x + 4$ . Show

Task 4

Is there a parabola through $(-1 \mid -1)$ , $(1 \mid 3)$ , $(2.5 \mid 6)$ ? Use the system of equations to explain.

Show solution

$\text{II} - \text{I}$ : $2b = 4$ , so $b = 2$ . $\text{III} - \text{II}$ : $5.25a + 1.5b = 3$ , so $a = 0$ , and then $c = 1$ . The points lie on the line $y = 2x + 1$ , so no parabola passes through them. Show

Task 5

Three positions of a thrown ball were measured (coordinates in m): $(-5 \mid 1)$ , $(-3 \mid 4)$ , $(3 \mid 1)$ . Find the parabola, the greatest height and the landing point.

Show solution

$(-5 \mid 1)$ and $(3 \mid 1)$ are at the same height, so the vertex lies halfway at $x = -1$ : $y = a(x + 1)^2 + y_S$ . Then $16a + y_S = 1$ and $4a + y_S = 4$ , so $a = -\frac{1}{4}$ and $y_S = 5$ : $f(x) = -\frac{1}{4}x^2 - \frac{1}{2}x + 4.75$ , greatest height $5\,\text{m}$ . Landing: $(x + 1)^2 = 20$ , so $x = -1 + \sqrt{20} \approx 3.47$ . Show

Task 6

A parabolic bridge arch is $10\,\text{m}$ wide at the bottom and $5\,\text{m}$ high. How high is it $3\,\text{m}$ from the centre?

Show solution

The points are $(-5 \mid 0)$ , $(0 \mid 5)$ and $(5 \mid 0)$ , which give $f(x) = -\frac{1}{5}x^2 + 5$ . Then $f(3) = 3.2$ , so the arch is $3.2\,\text{m}$ high there. Show

Ideas for the classroom

About grade 9, quadratic functions combined with systems of linear equations. How far systems with three unknowns are required in year 9 should be checked against your own curriculum and textbook. Good for consolidation and modelling (thrown ball, bridge arch).
The working uses II − I and III − I. Other orders are just as correct; comparing them with students’ solutions is worthwhile.
Points snap to a 0.5 grid, so the working stays exact. Non-terminating results are shown as reduced fractions.
Common mistakes: treating $x$ , $y$ as the unknowns, sign errors when squaring negative numbers, and believing that three points always determine a parabola.
Task mode: calc=0 together with Share → hide results hides the working, the equation and the vertex.