Quadratic Functions and Equations: math test solutions, Grade 11 – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

Sommaire

(a) \(f(x) = (x^2 - 8x + 16) - 16 + 11 = (x - 4)^2 - 5\). (2 pts)

(b) The vertex is \((4, -5)\), the axis is \(x = 4\) and the y-intercept is \(11\). (1 pt)

(a) Numbers with product \(-30\) and sum \(-7\): \(-10\) and \(3\). \(2x^2 - 10x + 3x - 15 = (2x + 3)(x - 5)\). (2 pts)

(b) \(4x^2 - 81 = (2x - 9)(2x + 9)\). (1 pt)

\(a = 2\), \(b = 3\), \(c = -7\): \(\Delta = 9 + 56 = 65\). (1 pt)

\(x = \dfrac{-3 \pm \sqrt{65}}{4}\). (2 pts)

\(x \approx 1.27\) or \(x \approx -2.77\). (1 pt)

(a) \(\Delta = 36 - 36 = 0\): one real solution (\(x = 3\)). (1 pt)

(b) \(\Delta = 4 - 48 = -44 < 0\): the solutions are complex, not real. (1 pt)

(a) \((x - 4)(x + 2) \leq 0\), so \(-2 \leq x \leq 4\), that is \([-2, 4]\). (1.5 pts)

(b) \(x^2 + 3x - 10 > 0\), so \((x + 5)(x - 2) > 0\), which gives \(x < -5\) or \(x > 2\). (1.5 pts)

(a) The vertex is at \(x = -\dfrac{2}{2 \cdot (-0.25)} = 4\), and \(h(4) = -4 + 8 + 1.75 = 5.75\). The maximum height is \(5.75\) feet. (1 pt)

(b) \(-0.25x^2 + 2x + 1.75 = 4.75\) gives \(x^2 - 8x + 12 = 0\), so \((x - 2)(x - 6) = 0\): at \(x = 2\) ft and \(x = 6\) ft. (1 pt)

(c) \(h(x) = 0\) gives \(x^2 - 8x - 7 = 0\), so \(x = 4 + \sqrt{23} \approx 8.8\) feet (the negative root is rejected). (1 pt)

Test yourself: quick challenge for Grade 11