GB-24-05-001 · Quadratic intercepts and turning points

The curve has equation

y=x^{2} - 16x + 15

(a) Find both x-intercepts.

(b) Find the coordinates of the turning point.

Part (a) Quadratic Graphs: Intercepts & Turning Point Booklet p. 12 & p. 14

Method

1) Factorise and set y = 0 to find x-intercepts

2) Complete the square (or average roots) to find turning point (−p, q)

Worked Example (GB-24-05-001 · Part (a))

Find both x-intercepts.

Factorise using 2×2 Grid (Booklet p. 5 & p. 12):

×	$x$	$-1$
$x$	$x^2$	$-x$
$-15$	$-15x$	$+15$

a \times c = 1 \times (15) = 15

$-1$	$-15$
$-3$	$-5$

c is positive so both signs are the same (− and −)...

-1 - 15 = -16

✔

\text{Brackets: } \mathbf{(x - 15)(x - 1)}

Set

y = 0 \Rightarrow (x - 15)(x - 1) = 0 \Rightarrow x = 15,\; x = 1

(1,0),\;(15,0)

Part (b) Quadratic Graphs: Intercepts & Turning Point Booklet p. 14

Method

1) Complete the square

2) Half the x coefficient inside the bracket

3) Subtract the added square

4) Read the turning point (−p, q)

Worked Example (GB-24-05-001 · Part (b))

Find the coordinates of the turning point.

Half the coefficient of x inside the bracket:

1[(x - 8)^2-(-8)^2] + 15

Simplify:

y=(x - 8)^2 - 49

The squared term is zero at

x=8

, giving turning point

(8,-49)

Since the coefficient of the square is positive, this is a minimum.

Ans =

(8,-49)