Egypt Preparatory 3 Mathematics, Week 2: Drawing the parabola: table of values, axis of symmetry, opening up or down, maximum or minimum value

The quadratic function

Week 2 · The quadratic function: standard form, vertex and axis of symmetry, vertex form, graph and maximum or minimum value · 3 lessons

Drawing the parabola: table of values, axis of symmetry, opening up or down, maximum or minimum value

Mathematics Prep 3 Term 1, Unit 2, Lesson 1

GoalDraw a parabola from a table of values, and say whether its vertex gives a maximum or a minimum value.

New words

opens up / opens down · a > 0: the arms go up; a < 0: the arms go downminimum value · the lowest y-value of the function, at the vertex when a > 0maximum value · the highest y-value of the function, at the vertex when a < 0table of values · a list of x-values with the matching f(x), used to plot points

Explanation

To draw a parabola: (1) find the vertex with x = −b/2a; (2) choose x-values on both sides of it, the same distance away, because equal distances give equal heights; (3) work out f(x) for each; (4) plot the points and join them with a smooth curve; (5) draw the axis of symmetry as a dashed line. The sign of a decides the shape. If a > 0 the parabola opens up, the vertex is the lowest point and f has a minimum value. If a < 0 it opens down, the vertex is the highest point and f has a maximum value. Be careful: the maximum or minimum VALUE is the y-coordinate of the vertex, not the x-coordinate.

Sign of a	Opens	Vertex is the	Value at the vertex
a > 0	up	lowest point	minimum
a < 0	down	highest point	maximum

Worked examples

Draw f(x) = −x² + 4x + 1. Here a = −1 < 0, so it opens down. The vertex has x = −4/(−2) = 2 and f(2) = −4 + 8 + 1 = 5. Take x from −1 to 5 (three steps each side of 2): f = −4, 1, 4, 5, 4, 1, −4. The maximum value is 5 (reached at x = 2).

x	0	1	2	3	4
f(x)	3	0	−1	0	3

Draw f(x) = x² − 4x + 3. Now a = 1 > 0, so it opens up. The vertex: x = 2, f(2) = 4 − 8 + 3 = −1. Table: x = 0, 1, 2, 3, 4 gives f = 3, 0, −1, 0, 3. The minimum value is −1.

Find the minimum value of f(x) = 2x² − 12x + 7. a = 2 > 0, so there is a minimum. x = 12/4 = 3 and f(3) = 18 − 36 + 7 = −11. The minimum value is −11 (not 3: 3 is where it happens).

Class activity

Draw it yourself. On squared paper plot f(x) = x² − 2x − 3 for x = −2 to 4 using your own table. Fold the paper along the axis x = 1. Do the two halves of the curve fit exactly on each other?

Practice

Complete the table for f(x) = x² − 2x − 3, x from −2 to 4, and name the vertex.

Does f(x) = −3x² + 6x + 2 have a maximum or a minimum? Find it.

Write the vertical line about which the parabola of f(x) = x² + 8x + 5 is symmetric.

Find the minimum value of f(x) = 3x² − 6x + 5.

(0, 5) is on the curve of f(x) = x² − 6x + 5. Use the axis x = 3 to find the mirror point with the same height.