Quadratic and Other Graphs

Grade 5

The quadratic graph — a parabola

The graph of y = ax² + bx + c is a parabola. If a > 0 it is U-shaped; if a < 0 it is arch-shaped. Key features to label: the y-intercept (at x = 0), the x-intercepts (roots — solve y = 0), and the vertex (minimum or maximum point, found by completing the square or using x = −b/(2a)).

Worked example — sketch a parabola

EXAMPLE

Sketch y = x² − 2x − 3, labelling the roots and turning point.

Roots: x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0, so x = 3 and x = −1. y-intercept: y = −3. Turning point: x = −(−2)/(2×1) = 1. y at x = 1: 1 − 2 − 3 = −4. Roots at x = −1 and x = 3; minimum at (1, −4); y-intercept at (0, −3).

Other graph shapes to recognise

You need to recognise: cubic (y = x³, S-shaped through origin). Reciprocal (y = k/x, two separate curves, never touching the axes). Exponential (y = aˣ, rapid growth through (0, 1), never negative).

Reading graphs

Intersection points of two graphs are the solutions to the simultaneous equations. Where y = f(x) meets y = g(x), the x-coordinate is the solution to f(x) = g(x). This links graphs directly to algebra.

Try this one

The diagram shows points \(R\) and \(S\) on a coordinate grid.

(a) Find the midpoint of \(RS\). [1]

(b) Find the equation of the line through \(R\) and \(S\). [3]

Nova's hint:

This question covers midpoints and linear graph equations.

Ask yourself: 'how do I find the midpoint of two points, and once I have the gradient, how do I use a point to find \(c\)?' For the midpoint, average the \(x\)-coordinates and average the \(y\)-coordinates separately.

Try to Solve it with Nova?