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Quiz: Quadratic Functions

Test your understanding of parabolas, the discriminant, solving techniques, and quadratic applications with these review questions.

1. Which direction does the parabola \(f(x) = -3x^2 + 4x - 1\) open?

  1. Upward
  2. Downward
  3. To the right
  4. To the left
Show Answer

The correct answer is B. The direction a parabola opens is determined by the sign of \(a\) (the coefficient of \(x^2\)). Here \(a = -3 < 0\), so the parabola opens downward, like an inverted bowl. Parabolas of the form \(y = ax^2 + bx + c\) always open vertically — never left or right. A downward parabola has a maximum point at its vertex.

Concept Tested: Parabola

2. What is the discriminant of \(2x^2 - 5x + 3 = 0\)?

  1. \(1\)
  2. \(-1\)
  3. \(49\)
  4. \(25\)
Show Answer

The correct answer is A. The discriminant is \(\Delta = b^2 - 4ac\). With \(a = 2\), \(b = -5\), \(c = 3\): \(\Delta = (-5)^2 - 4(2)(3) = 25 - 24 = 1\). Since \(\Delta > 0\), this equation has two distinct real roots. Be careful with signs — the \(b\) in the formula is squared, so its sign disappears in that term.

Concept Tested: Discriminant

3. A quadratic equation has \(\Delta < 0\). What does this tell you about its roots?

  1. Three real roots
  2. No real roots
  3. Two distinct real roots
  4. One repeated real root
Show Answer

The correct answer is B. When the discriminant is negative, the square root in the quadratic formula would be the square root of a negative number — which gives no real result. Graphically, the parabola doesn't intersect the \(x\)-axis at all. It hovers entirely above or entirely below. A quadratic can never have three roots, since the highest power is \(2\).

Concept Tested: No Real Roots

4. Find the vertex of the parabola \(f(x) = x^2 - 6x + 11\).

  1. \((6, 11)\)
  2. \((-3, 2)\)
  3. \((3, 2)\)
  4. \((3, -2)\)
Show Answer

The correct answer is C. The \(x\)-coordinate of the vertex is \(x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3\). The \(y\)-coordinate is \(f(3) = 9 - 18 + 11 = 2\). Alternatively, completing the square: \(f(x) = (x - 3)^2 + 2\), which shows the vertex is \((3, 2)\). Since \(a = 1 > 0\), this is a minimum point.

Concept Tested: Vertex of a Parabola

5. Factor \(x^2 - 7x + 10\).

  1. \((x + 2)(x + 5)\)
  2. \((x - 10)(x + 1)\)
  3. \((x - 1)(x - 10)\)
  4. \((x - 2)(x - 5)\)
Show Answer

The correct answer is D. Find two numbers that add to \(-7\) and multiply to \(10\). The pair \(-2\) and \(-5\) satisfy both: \(-2 + (-5) = -7\) and \((-2)(-5) = 10\). So \(x^2 - 7x + 10 = (x - 2)(x - 5)\). Option A would give \(x^2 + 7x + 10\), and option C gives \(x^2 - 11x + 10\), neither of which matches.

Concept Tested: Factoring Quadratics

6. Solve \(x^2 + 2x - 8 = 0\).

  1. \(x = 2\) or \(x = 4\)
  2. \(x = -2\) or \(x = 4\)
  3. \(x = 2\) or \(x = -4\)
  4. \(x = -2\) or \(x = -4\)
Show Answer

The correct answer is C. Factor: we need two numbers that add to \(2\) and multiply to \(-8\). The pair \(4\) and \(-2\) works: \(4 + (-2) = 2\) and \((4)(-2) = -8\). So \((x + 4)(x - 2) = 0\), giving \(x = -4\) or \(x = 2\). Verify: \((2)^2 + 2(2) - 8 = 4 + 4 - 8 = 0\). ✓

Concept Tested: Solving Quadratic Equations

7. Write \(f(x) = x^2 + 8x + 10\) in vertex form by completing the square.

  1. \((x + 4)^2 - 6\)
  2. \((x + 4)^2 + 26\)
  3. \((x - 4)^2 - 6\)
  4. \((x + 8)^2 - 54\)
Show Answer

The correct answer is A. Take half of the coefficient of \(x\): \(\frac{8}{2} = 4\). Square it: \(4^2 = 16\). Rewrite: \(f(x) = (x^2 + 8x + 16) - 16 + 10 = (x + 4)^2 - 6\). The vertex is \((-4, -6)\). Verify by expanding: \((x + 4)^2 - 6 = x^2 + 8x + 16 - 6 = x^2 + 8x + 10\). ✓

Concept Tested: Completing the Square

8. A ball is thrown upward with height function \(h(t) = -5t^2 + 30t + 2\). What is the maximum height reached?

  1. \(32\) m
  2. \(30\) m
  3. \(47\) m
  4. \(45\) m
Show Answer

The correct answer is C. The maximum occurs at the vertex of the parabola. The \(t\)-value is \(t = -\frac{30}{2(-5)} = 3\) seconds. Then \(h(3) = -5(9) + 30(3) + 2 = -45 + 90 + 2 = 47\) m. A common mistake is to plug in \(t = 0\) (which gives only the initial height) or to forget to square \(t\) correctly.

Concept Tested: Projectile Motion Model

9. Solve the inequality \(x^2 - 4 > 0\).

  1. \(-2 < x < 2\)
  2. \(x < -2\) or \(x > 2\)
  3. \(x > 2\)
  4. \(-2 \leq x \leq 2\)
Show Answer

The correct answer is B. First solve \(x^2 - 4 = 0\) to find the roots: \(x = \pm 2\). The parabola \(y = x^2 - 4\) opens upward, so it is above the \(x\)-axis (positive) outside the roots. Testing intervals confirms: at \(x = 0\), \(0 - 4 = -4 < 0\); at \(x = 3\), \(9 - 4 = 5 > 0\). The solution is \(x < -2\) or \(x > 2\), i.e., \((-\infty, -2) \cup (2, \infty)\).

Concept Tested: Quadratic Inequalities

10. A quadratic function is written as \(f(x) = 2(x - 3)(x + 1)\). What are the roots?

  1. \(x = 3\) and \(x = -1\)
  2. \(x = -3\) and \(x = 1\)
  3. \(x = 3\) and \(x = 1\)
  4. \(x = 2\) and \(x = -1\)
Show Answer

The correct answer is A. Factored form \(f(x) = a(x - p)(x - q)\) immediately shows the roots are \(x = p\) and \(x = q\). Here, set each factor equal to zero: \(x - 3 = 0\) gives \(x = 3\), and \(x + 1 = 0\) gives \(x = -1\). The leading coefficient \(2\) affects the width and direction but not the root locations.

Concept Tested: Factored Form Quadratic