Purpose: Self-check factoring, completing the square, the quadratic formula, and applications.
Score: 0 / 12
Question 1
The vertex of \( y = -2(x - 3)^2 + 5 \) is:
Solution:
Vertex form \( y = a(x-h)^2 + k \) gives vertex \( (h, k) = (3, 5) \).
Question 2
Factor: \( x^2 - 11x + 24 \).
Solution:
Need two numbers whose sum is \( -11 \) and product is \( 24 \): \( -8 \) and \( -3 \). So \( (x-8)(x-3) \).
Question 3
Factor completely: \( 6x^2 + 7x - 5 \).
Solution:
Decomposition: need \( ac = 6 \cdot (-5) = -30 \), \( b = 7 \). Numbers: 10, -3. \( 6x^2 + 10x - 3x - 5 = 2x(3x + 5) - (3x + 5) = (3x + 5)(2x - 1) \).
Question 4
Factor \( 9x^2 - 16 \) as a difference of squares.
Solution:
\( 9x^2 - 16 = (3x)^2 - 4^2 = (3x - 4)(3x + 4) \).
Question 5
Convert \( y = x^2 - 6x + 11 \) to vertex form. The vertex \( (h, k) \) is at \( h = \) ?
Solution:
\( y = (x^2 - 6x + 9) - 9 + 11 = (x - 3)^2 + 2 \). So \( h = 3 \), \( k = 2 \).
Question 6
For \( x^2 + 4x + 7 = 0 \), the discriminant \( \Delta = b^2 - 4ac \) equals:
Solution:
\( \Delta = 4^2 - 4(1)(7) = 16 - 28 = -12 \). Since \( \Delta < 0 \), there are no real roots.
Question 7
Solve \( x^2 - 5x + 6 = 0 \). The smaller root is:
Solution:
\( (x - 2)(x - 3) = 0 \), so \( x = 2 \) or \( x = 3 \). Smaller is 2.
Question 8
Using the quadratic formula, the sum of the roots of \( x^2 - 6x + 4 = 0 \) is:
Solution:
By Vieta or formula: roots are \( 3 \pm \sqrt{5} \). Their sum is \( 6 \).
Question 9
A ball thrown straight up has \( h(t) = -5t^2 + 30t + 2 \) (in metres). The time at maximum height is \( t = \) ?
Solution:
\( t = -\frac{b}{2a} = -\frac{30}{2(-5)} = 3 \) seconds.
Question 10
From Q9, the maximum height (m) is:
Solution:
\( h(3) = -5(9) + 30(3) + 2 = -45 + 90 + 2 = 47 \) m.
Question 11
A rectangle's perimeter is 40 m. Its area in terms of width \( w \) is \( A(w) = w(20 - w) \). The maximum area (m²) is:
Solution:
Vertex at \( w = 10 \), giving \( A = 10 \cdot 10 = 100 \) m². Always a square.
Question 12
A profit function is \( P(x) = -x^2 + 14x - 24 \) (thousands of dollars). The break-even values of \( x \) (where \( P = 0 \)) are 2 and:
Solution:
\( -x^2 + 14x - 24 = 0 \Rightarrow x^2 - 14x + 24 = 0 \Rightarrow (x - 2)(x - 12) = 0 \). So \( x = 2 \) or \( x = 12 \).