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RV-20-001 · Quadratic graph: factorisation, vertex and intersection

The curve has equation y=x2−9y=x^{2} - 9.
-5-4-3-2-1012345-12-9-6-33691215xy

(a) Factorise the expression for y and write down the x-intercepts.

(b) Find the turning point. Explain whether it is a maximum or minimum.

Part (a) Quadratic Graphs: Factorising, Roots & Completed-Square Vertex Booklet p. 5, p. 12 & p. 14
Method
1) Factorise quadratic and set y = 0 for x-intercepts
2) Complete the square y = (x − h)² + k to state turning point (h, k) and explain minimum
Worked Example (RV-20-001 · Part (a))
Factorise the expression for y and write down the x-intercepts.
Booklet p. 6 (DOTS): a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)
x2−9=(x+3)(x−3)x^2 - 9 = \mathbf{(x + 3)(x - 3)}
Set y=0⇒x=−3y = 0 \Rightarrow x = -3 and x=3⇒(−3,0)x = 3 \Rightarrow (-3, 0) and (3,0)(3, 0)
(x+3)(x−3)(x + 3)(x - 3); (-3, 0), (3, 0).
Part (b) Quadratic Graphs: Factorising, Roots & Completed-Square Vertex 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 (RV-20-001 · Part (b))
Find the turning point. Explain whether it is a maximum or minimum.
1)
Half the coefficient of x inside the bracket: 1[(x+0)2−(0)2]−91[(x + 0)^2-(0)^2] - 9.
2)
Simplify: y=(x+0)2−9y=(x + 0)^2 - 9.
3)
The squared term is zero at x=0x=0, giving turning point (0,−9)(0,-9).
4)
Since the coefficient of the square is positive, this is a minimum.
Ans = (0, -9), minimum.