Sketching Quadratic Graphs

Sketching a quadratic means showing its shape and key features in the right places, not plotting it point by point. Three features do the work: which way it opens, where it crosses the axes, and where its vertex sits.

The three features

• Opening direction: the sign of the $x^2$ coefficient. Positive opens upward ($\cup$, a minimum); negative opens downward ($\cap$, a maximum).
• $x$-intercepts: solve $y = 0$ (factorise, or use the formula). These are where the curve crosses the $x$-axis. If the discriminant is negative, there are none, the curve doesn’t touch the axis.
• $y$-intercept: set $x = 0$, which is just the constant term.
• Vertex: from completing the square, or note the vertex lies midway between the roots.

A worked sketch

Sketch $y = x^2 - 2x - 8$. Opens upward (positive $x^2$). $x$-intercepts: $(x-4)(x+2)=0 \to x = 4, -2$. $y$-intercept: $-8$. Vertex: $(x-1)^2 - 9 \to (1, -9)$. So: a $\cup$-parabola crossing the $x$-axis at $-2$ and $4$, the $y$-axis at $-8$, minimum at $(1, -9)$.

Label these coordinates on the sketch, the marks are for the features being correct and labelled, not for a neat freehand curve.

When there are no real roots

If $b^2 - 4ac < 0$, the parabola sits entirely above the $x$-axis (upward) or below it (downward), touching neither. Show it floating clear of the axis, with only the $y$-intercept and vertex marked.

Common mistakes

• Getting the opening direction wrong (check the sign of the $x^2$ term).
• Swapping the $x$- and $y$-intercepts.
• Misreading the vertex $x$-coordinate from the completed square (from $(x-1)^2$ it’s $+1$).

Full topic context: Quadratic Functions notes, and the worked sketching example.