Quadratic Graphs

Video lesson5 worked examples

Theory

The graph of a quadratic function is a parabola. You should be able to read its key features straight from the equation, and — going the other way — work out the equation from a graph.

Reading Features from the Equation

Shape & nature: a positive $x^2$ term gives a “smile” with a minimum; a negative $x^2$ term gives a “frown” with a maximum.
$y$ -intercept: the constant term $c$ — the graph crosses the $y$ -axis at $(0, c)$ .
Roots / zeros: where the graph crosses the $x$ -axis — found by solving $y = 0$ (factorise).
Turning point & axis of symmetry: read directly from completed-square (vertex) form $y = a(x + p)^2 + q$ , giving $(-p, q)$ and $x = -p$ .

Determining the Equation from a Graph

$y = kx^2$ : turning point at the origin. Substitute a known point to find $k$ .
Turning point form $y = k(x - a)^2 + b$ : read $(a, b)$ from the turning point, then use another point to find $k$ .
Root form $y = k(x - p)(x - q)$ : read the roots $p$ and $q$ , then use another point to find $k$ .

The Golden Rule: match the information you are given to the right form — origin turning point means $y = kx^2$ ; a turning point elsewhere means vertex form; visible roots mean root form. Then substitute one more point to pin down $k$ .

⚠️ Common Examiner Traps

Sign of the roots in root form: a root at $x = 3$ gives a factor $(x - 3)$ ; a root at $x = -3$ gives $(x + 3)$ .
Turning-point sign: vertex form $(x - a)^2$ has its turning point at $x = +a$ , not $-a$ .
Forgetting $k$ : the shape can be stretched, so you must find $k$ from an extra point — don't assume it is 1.
Reading the y-intercept as a root: the $y$ -intercept is where $x = 0$ ; the roots are where $y = 0$ .

Worked examples

Example 1

Identifying Features

For the parabola $y = x^2 - 2x - 8$ , state the shape, the $y$ -intercept, and the roots.

Step 1: The $x^2$ term is positive, so the parabola is a “smile” with a minimum.

Step 2: The constant term is $-8$ , so the $y$ -intercept is $(0, -8)$ .

Step 3: For the roots, set $y = 0$ and factorise: $x^2 - 2x - 8 = (x - 4)(x + 2) = 0$ .

Answer: minimum parabola, $y$ -intercept $(0, -8)$ , roots at $(4, 0)$ and $(-2, 0)$ .

Example 2

Reading Vertex Form

A parabola has equation $y = 2(x - 4)^2 + 3$ . State the coordinates of its turning point, its nature, and the equation of its axis of symmetry.

Step 1: Compare with $y = a(x + p)^2 + q$ . Here $p = -4$ and $q = 3$ , so the turning point $(-p, q)$ is $(4, 3)$ .

Step 2: The coefficient $a = 2$ is positive, so the turning point is a minimum.

Answer: minimum turning point $(4, 3)$ , axis of symmetry $x = 4$ .

Example 3

Equation of the form $y = kx^2$

A parabola with equation $y = kx^2$ passes through the point $(3, 18)$ . Find the value of $k$ and state the equation.

Step 1: The turning point is at the origin, so the form is $y = kx^2$ . Substitute the given point $x = 3,\ y = 18$ :

18 = k \times 3^2 = 9k

Step 2: Solve for $k$ : $k = 2$ .

Answer: $k = 2$ , so the equation is $y = 2x^2$ .

Example 4

🎯 Exam-style (equation from the turning point)

A parabola has turning point $(3, -4)$ and passes through the point $(5, 8)$ . Determine its equation.

Step 1: A turning point away from the origin means vertex form $y = k(x - a)^2 + b$ with $(a, b) = (3, -4)$ :

y = k(x - 3)^2 - 4

Step 2: Use the other point $(5, 8)$ to find $k$ . Substitute $x = 5,\ y = 8$ :

8 = k(5 - 3)^2 - 4 \implies 8 = 4k - 4 \implies 4k = 12 \implies k = 3

Answer: $y = 3(x - 3)^2 - 4$ .

Example 5

Equation from the roots

A parabola cuts the $x$ -axis at $(-1, 0)$ and $(4, 0)$ , and passes through $(0, -8)$ . Find its equation.

Step 1: With the roots visible, use root form $y = k(x - p)(x - q)$ . Roots at $-1$ and $4$ give factors $(x + 1)$ and $(x - 4)$ :

y = k(x + 1)(x - 4)

Step 2: Use the point $(0, -8)$ to find $k$ :

-8 = k(0 + 1)(0 - 4) = -4k \implies k = 2

Answer: $y = 2(x + 1)(x - 4)$ .

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22 Quadratics questions with answers and video solutions