Quadratic functions - Answer key

Fluency

$a = 2$ , $b = -4$ , $c = 1$ . Since $a = 2 > 0$ , the parabola opens upward.
$y$ -intercept: $(0, -5)$ .
$(x + 3)(x + 4)$ .
$(x - 4)(x - 5)$ .
$(x + 5)(x - 3)$ .
$x = 3$ or $x = -7$ .
$(x - 2)(x - 3) = 0$ : $x = 2$ or $x = 3$ .
$(x + 4)(x - 3) = 0$ : $x = -4$ or $x = 3$ .
$(x - 4)(x + 4) = 0$ : $x = 4$ or $x = -4$ .
$3x(x - 4) = 0$ : $x = 0$ or $x = 4$ .

Reasoning

$x = -\dfrac{-4}{2(1)} = 2$ . Turning point: $y = 4 - 8 + 3 = -1$ . Turning point is $(2, -1)$ .
$x$ -intercepts: $x^2 - 2x - 8 = (x - 4)(x + 2) = 0$ , so $x = 4$ or $x = -2$ . Axis of symmetry: $x = 1$ . Turning point: $y = 1 - 2 - 8 = -9$ , so $(1, -9)$ . $y$ -intercept: $(0, -8)$ . Parabola opens upward.
Axis of symmetry: $x = \dfrac{-3 + 1}{2} = -1$ . Turning point: $y = 1 - 2 - 3 = -4$ , so $(-1, -4)$ .
$x^2 - 7x + 10 = 0$ . $(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
$2(x^2 - 7x + 10) = 0$ . $2(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
$x^2 = -4$ has no real solution because a square is never negative. The parabola $y = x^2 + 4$ has its turning point at $(0, 4)$ , entirely above the $x$ -axis, so it never crosses it.
$(x + 3)(x - 1) = 21$ . Expand: $x^2 + 2x - 3 = 21$ , so $x^2 + 2x - 24 = 0$ . $(x + 6)(x - 4) = 0$ : $x = -6$ or $x = 4$ . Since $x - 1 > 0$ , we need $x > 1$ , so $x = 4$ . The rectangle is $7$ cm by $3$ cm.
Let the integers be $n$ and $n + 1$ . $n(n + 1) = 72$ , so $n^2 + n - 72 = 0$ . $(n + 9)(n - 8) = 0$ : $n = 8$ or $n = -9$ . The consecutive integers are $8$ and $9$ (or $-9$ and $-8$ ).

Reasoning

In $y = (x - p)(x - q)$ , the $x$ -intercepts are $x = p$ and $x = q$ (by the null factor law). The axis of symmetry is $x = \dfrac{p + q}{2}$ (midpoint of the intercepts).
Ground: $24t - 4t^2 = 0$ , $4t(6 - t) = 0$ , so $t = 0$ or $t = 6$ s. Axis of symmetry: $t = 3$ . Max height: $h = 24(3) - 4(9) = 72 - 36 = 36$ m.
Let the numbers be $x$ and $12 - x$ . Product: $x(12 - x) = 32$ , so $12x - x^2 = 32$ , $x^2 - 12x + 32 = 0$ . $(x - 4)(x - 8) = 0$ : $x = 4$ or $x = 8$ . The numbers are $4$ and $8$ .
One $x$ -intercept means the discriminant is zero: $b^2 - 4(1)(9) = 0$ , so $b^2 = 36$ , $b = 6$ or $b = -6$ .

Reasoning

Width $= x$ , length $= 60 - 2x$ . Area $= x(60 - 2x) = 60x - 2x^2$ . This is a downward parabola. Axis of symmetry: $x = -\dfrac{60}{2(-2)} = 15$ . Maximum area: $A = 60(15) - 2(225) = 900 - 450 = 450$ m $^2$ . Dimensions: $15$ m wide, $30$ m long.
Substitute $(1, 2)$ : $2 = 1 - 6 + k$ , so $k = 7$ . Equation: $y = x^2 - 6x + 7$ . Turning point: $x = 3$ , $y = 9 - 18 + 7 = -2$ , so $(3, -2)$ . $x$ -intercepts: $x^2 - 6x + 7 = 0$ . Using the quadratic formula (or completing the square): $x = \dfrac{6 \pm \sqrt{36 - 28}}{2} = \dfrac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$ .
$(x - p)(x - q) = x^2 - (p+q)x + pq$ . Comparing with $x^2 + bx + c$ : $-(p+q) = b$ so $p + q = -b$ , and $pq = c$ .
Let the numbers be $n$ and $n + 2$ . $n^2 + (n+2)^2 = 130$ . $2n^2 + 4n + 4 = 130$ . $n^2 + 2n - 63 = 0$ . $(n + 9)(n - 7) = 0$ : $n = 7$ (positive). The numbers are $7$ and $9$ .

Year 9 core - answers