IB Mathematics: Analysis and Approaches Quiz: Quadratic Functions
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Quadratic FunctionsQuestion 1 of 20

The parabola y=x2−2kx+ky = x^2 - 2kx + k lies entirely above the x-axis. Which condition must kk satisfy?

k<0k < 0
k>1k > 1
0<k<10 < k < 1
k<0k < 0 or k>1k > 1
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Quadratic Functions

Practice Quadratic Functions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Quadratic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The parabola y=x2−2kx+ky = x^2 - 2kx + k lies entirely above the x-axis. Which condition must kk satisfy?

  1. k<0k < 0
  2. k>1k > 1
  3. 0<k<10 < k < 1 (correct answer)
  4. k<0k < 0 or k>1k > 1
Explanation: For a parabola to lie entirely above the x-axis, two conditions must be met:
  1. The parabola must open upwards. The coefficient of x2x^2 is a=1a=1, which is positive, so this condition is met.
  2. The parabola must not intersect the x-axis, which means the quadratic equation x2−2kx+k=0x^2 - 2kx + k = 0 has no real roots. This happens when the discriminant Δ=b2−4ac\Delta = b^2 - 4ac is negative.
Here, a=1,b=−2k,c=ka=1, b=-2k, c=k. Δ=(−2k)2−4(1)(k)<0\Delta = (-2k)^2 - 4(1)(k) < 0 4k2−4k<04k^2 - 4k < 0 Divide by 4: k2−k<0k^2 - k < 0 Factor the expression: k(k−1)<0k(k-1) < 0 The roots of k(k−1)=0k(k-1)=0 are k=0k=0 and k=1k=1. The expression k2−kk^2-k represents an upward-opening parabola in kk, so it is negative between its roots. Therefore, 0<k<10 < k < 1. Distractor D is the solution for Δ>0\Delta > 0, which corresponds to the parabola having two distinct x-intercepts.

Question 2

The equation kx2−4x+(k−3)=0kx^2 - 4x + (k-3) = 0 has two distinct real roots. Which of the following describes the set of possible values for kk?

  1. −1<k<4-1 < k < 4
  2. k<−1k < -1 or k>4k > 4
  3. −1<k<4,k≠0-1 < k < 4, k \neq 0 (correct answer)
  4. k=−1k=-1 or k=4k=4
Explanation: For a quadratic equation to have two distinct real roots, the discriminant Δ=b2−4ac\Delta = b^2 - 4ac must be positive. Here, a=k,b=−4,c=k−3a=k, b=-4, c=k-3. Δ=(−4)2−4(k)(k−3)>0\Delta = (-4)^2 - 4(k)(k-3) > 0 16−4k2+12k>016 - 4k^2 + 12k > 0 Dividing by -4 and reversing the inequality sign gives: k2−3k−4<0k^2 - 3k - 4 < 0 Factoring the quadratic expression: (k−4)(k+1)<0(k-4)(k+1) < 0 The roots of k2−3k−4=0k^2 - 3k - 4 = 0 are k=−1k=-1 and k=4k=4. Since the parabola opens upwards, the expression is negative between the roots, so −1<k<4-1 < k < 4. Additionally, for the equation to be quadratic, the leading coefficient kk cannot be zero. Therefore, the complete set of values is −1<k<4,k≠0-1 < k < 4, k \neq 0.

Question 3

The parabola y=x2−2kx+ky = x^2 - 2kx + k lies entirely above the x-axis. Which condition must kk satisfy?

  1. k<0k < 0
  2. k>1k > 1
  3. 0<k<10 < k < 1 (correct answer)
  4. k<0k < 0 or k>1k > 1
Explanation: For a parabola to lie entirely above the x-axis, two conditions must be met:
  1. The parabola must open upwards. The coefficient of x2x^2 is a=1a=1, which is positive, so this condition is met.
  2. The parabola must not intersect the x-axis, which means the quadratic equation x2−2kx+k=0x^2 - 2kx + k = 0 has no real roots. This happens when the discriminant Δ=b2−4ac\Delta = b^2 - 4ac is negative.
Here, a=1,b=−2k,c=ka=1, b=-2k, c=k. Δ=(−2k)2−4(1)(k)<0\Delta = (-2k)^2 - 4(1)(k) < 0 4k2−4k<04k^2 - 4k < 0 Divide by 4: k2−k<0k^2 - k < 0 Factor the expression: k(k−1)<0k(k-1) < 0 The roots of k(k−1)=0k(k-1)=0 are k=0k=0 and k=1k=1. The expression k2−kk^2-k represents an upward-opening parabola in kk, so it is negative between its roots. Therefore, 0<k<10 < k < 1. Distractor D is the solution for Δ>0\Delta > 0, which corresponds to the parabola having two distinct x-intercepts.

Question 4

Find the set of values of xx for which 3x+10<x23x + 10 < x^2.

  1. −2<x<5-2 < x < 5
  2. x<−5x < -5 or x>2x > 2
  3. x<−2x < -2 or x>5x > 5 (correct answer)
  4. x>5x > 5
Explanation: First, rearrange the inequality to the standard quadratic form: x2−3x−10>0x^2 - 3x - 10 > 0 Next, find the roots of the corresponding quadratic equation x2−3x−10=0x^2 - 3x - 10 = 0 by factoring: (x−5)(x+2)=0(x-5)(x+2) = 0 The roots are x=5x=5 and x=−2x=-2. These roots are the critical values where the expression equals zero. We need to find where the expression is greater than zero. The graph of y=x2−3x−10y = x^2 - 3x - 10 is an upward-opening parabola. It will be above the x-axis (i.e., positive) when xx is outside the roots. Therefore, the solution is x<−2x < -2 or x>5x > 5. Distractor A is the solution to x2−3x−10<0x^2 - 3x - 10 < 0. Distractor B results from incorrectly factoring the quadratic as (x+5)(x−2)(x+5)(x-2).

Question 5

For which values of mm does the equation x2+mx+m+3=0x^2 + mx + m + 3 = 0 have no real roots?

  1. m<−2m < -2 or m>6m > 6
  2. −6<m<2-6 < m < 2
  3. −2<m<6-2 < m < 6 (correct answer)
  4. −3<m<4-3 < m < 4
Explanation: A quadratic equation has no real roots if its discriminant Δ=b2−4ac\Delta = b^2 - 4ac is less than zero. For the equation x2+mx+m+3=0x^2 + mx + m + 3 = 0, we have a=1,b=m,c=m+3a=1, b=m, c=m+3. We set Δ<0\Delta < 0: Δ=m2−4(1)(m+3)<0\Delta = m^2 - 4(1)(m+3) < 0 m2−4m−12<0m^2 - 4m - 12 < 0 To solve this quadratic inequality, we first find the roots of m2−4m−12=0m^2 - 4m - 12 = 0. Factoring the expression: (m−6)(m+2)=0(m-6)(m+2) = 0 The roots are m=6m=6 and m=−2m=-2. The graph of y=m2−4m−12y = m^2 - 4m - 12 is an upward-opening parabola. The expression is negative between the roots. Therefore, the solution is −2<m<6-2 < m < 6. Distractor A is the solution for Δ>0\Delta > 0 (two distinct real roots). Distractor B results from incorrectly factoring the quadratic as (m+6)(m−2)(m+6)(m-2).

Question 6

Find the sum of the solutions to the equation 4x−5⋅2x+4=04^x - 5 \cdot 2^x + 4 = 0.

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 5
Explanation: The equation involves exponential terms. We can rewrite 4x4^x as (22)x=(2x)2(2^2)^x = (2^x)^2. The equation becomes: (2x)2−5(2x)+4=0(2^x)^2 - 5(2^x) + 4 = 0 This is a quadratic equation in terms of 2x2^x. Let u=2xu = 2^x. The equation transforms to: u2−5u+4=0u^2 - 5u + 4 = 0 We can solve this quadratic for uu by factoring: (u−1)(u−4)=0(u-1)(u-4) = 0 This gives two possible values for uu: u=1u=1 or u=4u=4. Now, we substitute back u=2xu = 2^x to find the values of xx: Case 1: 2x=12^x = 1. This implies x=0x=0. Case 2: 2x=42^x = 4. This implies x=2x=2. The solutions to the original equation are x=0x=0 and x=2x=2. The sum of the solutions is 0+2=20 + 2 = 2. Distractor D is the sum of the roots of the quadratic in uu, not xx. Distractors A and C are the individual solutions for xx.

Question 7

The equation kx2−4x+(k−3)=0kx^2 - 4x + (k-3) = 0 has two distinct real roots. Which of the following describes the set of possible values for kk?

  1. −1<k<4-1 < k < 4
  2. k<−1k < -1 or k>4k > 4
  3. −1<k<4,k≠0-1 < k < 4, k \neq 0 (correct answer)
  4. k=−1k=-1 or k=4k=4
Explanation: For a quadratic equation to have two distinct real roots, the discriminant Δ=b2−4ac\Delta = b^2 - 4ac must be positive. Here, a=k,b=−4,c=k−3a=k, b=-4, c=k-3. Δ=(−4)2−4(k)(k−3)>0\Delta = (-4)^2 - 4(k)(k-3) > 0 16−4k2+12k>016 - 4k^2 + 12k > 0 Dividing by -4 and reversing the inequality sign gives: k2−3k−4<0k^2 - 3k - 4 < 0 Factoring the quadratic expression: (k−4)(k+1)<0(k-4)(k+1) < 0 The roots of k2−3k−4=0k^2 - 3k - 4 = 0 are k=−1k=-1 and k=4k=4. Since the parabola opens upwards, the expression is negative between the roots, so −1<k<4-1 < k < 4. Additionally, for the equation to be quadratic, the leading coefficient kk cannot be zero. Therefore, the complete set of values is −1<k<4,k≠0-1 < k < 4, k \neq 0.

Question 8

The function f(x)=x2−6x+13f(x) = x^2 - 6x + 13 is defined for x≥3x \ge 3. Find the inverse function f−1(x)f^{-1}(x).

  1. f−1(x)=3−x−4f^{-1}(x) = 3 - \sqrt{x-4}
  2. f−1(x)=−3+x−4f^{-1}(x) = -3 + \sqrt{x-4}
  3. f−1(x)=3+x+4f^{-1}(x) = 3 + \sqrt{x+4}
  4. f−1(x)=3+x−4f^{-1}(x) = 3 + \sqrt{x-4} (correct answer)
Explanation: When finding the inverse of a function with a restricted domain, you need to complete the square, solve for the inverse algebraically, and carefully consider which branch to take based on the domain restriction. Start by completing the square for f(x)=x2−6x+13f(x) = x^2 - 6x + 13. Take half of the coefficient of xx (which is −6-6), square it: (−3)2=9(-3)^2 = 9. This gives us f(x)=(x−3)2+4f(x) = (x-3)^2 + 4. To find the inverse, set y=(x−3)2+4y = (x-3)^2 + 4 and solve for xx. Subtract 4: y−4=(x−3)2y-4 = (x-3)^2. Take the square root: ±y−4=x−3\pm\sqrt{y-4} = x-3. This means x=3±y−4x = 3 \pm \sqrt{y-4}. Here's the crucial step: since the original function is restricted to x≥3x \geq 3, you need the positive branch. When x≥3x \geq 3, we have x−3≥0x-3 \geq 0, so x=3+y−4x = 3 + \sqrt{y-4}. Therefore, f−1(x)=3+x−4f^{-1}(x) = 3 + \sqrt{x-4}. Option A gives 3−x−43 - \sqrt{x-4}, which would correspond to the left branch of the parabola (x<3x < 3), contradicting our domain restriction. Option B has −3+x−4-3 + \sqrt{x-4}, which shifts the vertex incorrectly. Option C has x+4\sqrt{x+4} instead of x−4\sqrt{x-4}, indicating an error in the completing the square process. Study tip: Always check that your inverse function's range matches the original function's domain. Domain restrictions in quadratic functions typically tell you which branch of the square root to choose when finding the inverse.

Question 9

The function f(x)=2x2−8x+11f(x) = 2x^2 - 8x + 11 can be written in the vertex form f(x)=a(x−h)2+kf(x) = a(x-h)^2 + k. What is the minimum value of this function?

  1. 2
  2. 3 (correct answer)
  3. 11
  4. 19
Explanation: The minimum value of a parabola that opens upwards (since a=2>0a=2>0) is the y-coordinate of its vertex, kk. We can find the vertex by completing the square or using the formula x=−b/(2a)x = -b/(2a). Method 1: Completing the square. f(x)=2(x2−4x)+11f(x) = 2(x^2 - 4x) + 11 f(x)=2(x2−4x+4−4)+11f(x) = 2(x^2 - 4x + 4 - 4) + 11 f(x)=2((x−2)2−4)+11f(x) = 2((x-2)^2 - 4) + 11 f(x)=2(x−2)2−8+11f(x) = 2(x-2)^2 - 8 + 11 f(x)=2(x−2)2+3f(x) = 2(x-2)^2 + 3 The vertex is at (2,3)(2, 3). The minimum value is 3. Method 2: Using the vertex formula. The x-coordinate of the vertex is h=−b2a=−−82(2)=84=2h = -\frac{b}{2a} = -\frac{-8}{2(2)} = \frac{8}{4} = 2. The minimum value is the y-coordinate, k=f(h)=f(2)k = f(h) = f(2). f(2)=2(2)2−8(2)+11=2(4)−16+11=8−16+11=3f(2) = 2(2)^2 - 8(2) + 11 = 2(4) - 16 + 11 = 8 - 16 + 11 = 3. Distractor A is the x-coordinate of the vertex. Distractor C is the y-intercept. Distractor D results from a sign error in completing the square (adding 8 instead of subtracting).

Question 10

Find the sum of the solutions to the equation 4x−5⋅2x+4=04^x - 5 \cdot 2^x + 4 = 0.

  1. 1
  2. 2 (correct answer)
  3. 4
  4. 5
Explanation: The equation involves exponential terms. We can rewrite 4x4^x as (22)x=(2x)2(2^2)^x = (2^x)^2. The equation becomes: (2x)2−5(2x)+4=0(2^x)^2 - 5(2^x) + 4 = 0 This is a quadratic equation in terms of 2x2^x. Let u=2xu = 2^x. The equation transforms to: u2−5u+4=0u^2 - 5u + 4 = 0 We can solve this quadratic for uu by factoring: (u−1)(u−4)=0(u-1)(u-4) = 0 This gives two possible values for uu: u=1u=1 or u=4u=4. Now, we substitute back u=2xu = 2^x to find the values of xx: Case 1: 2x=12^x = 1. This implies x=0x=0. Case 2: 2x=42^x = 4. This implies x=2x=2. The solutions to the original equation are x=0x=0 and x=2x=2. The sum of the solutions is 0+2=20 + 2 = 2. Distractor D is the sum of the roots of the quadratic in uu, not xx. Distractors A and C are the individual solutions for xx.

Question 11

For which values of mm does the equation x2+mx+m+3=0x^2 + mx + m + 3 = 0 have no real roots?

  1. m<−2m < -2 or m>6m > 6
  2. −6<m<2-6 < m < 2
  3. −2<m<6-2 < m < 6 (correct answer)
  4. −3<m<4-3 < m < 4
Explanation: A quadratic equation has no real roots if its discriminant Δ=b2−4ac\Delta = b^2 - 4ac is less than zero. For the equation x2+mx+m+3=0x^2 + mx + m + 3 = 0, we have a=1,b=m,c=m+3a=1, b=m, c=m+3. We set Δ<0\Delta < 0: Δ=m2−4(1)(m+3)<0\Delta = m^2 - 4(1)(m+3) < 0 m2−4m−12<0m^2 - 4m - 12 < 0 To solve this quadratic inequality, we first find the roots of m2−4m−12=0m^2 - 4m - 12 = 0. Factoring the expression: (m−6)(m+2)=0(m-6)(m+2) = 0 The roots are m=6m=6 and m=−2m=-2. The graph of y=m2−4m−12y = m^2 - 4m - 12 is an upward-opening parabola. The expression is negative between the roots. Therefore, the solution is −2<m<6-2 < m < 6. Distractor A is the solution for Δ>0\Delta > 0 (two distinct real roots). Distractor B results from incorrectly factoring the quadratic as (m+6)(m−2)(m+6)(m-2).

Question 12

The quadratic equation 3x2−9x+2=03x^2 - 9x + 2 = 0 has roots α\alpha and β\beta. Find the value of α2+β2\alpha^2 + \beta^2.

  1. 5/35/3
  2. 23/323/3 (correct answer)
  3. 99
  4. 31/331/3
Explanation: For the quadratic equation ax2+bx+c=0ax^2+bx+c=0, the sum of roots is α+β=−b/a\alpha + \beta = -b/a and the product of roots is αβ=c/a\alpha\beta = c/a. For 3x2−9x+2=03x^2 - 9x + 2 = 0, we have a=3,b=−9,c=2a=3, b=-9, c=2. Sum of roots: α+β=−(−9)/3=3\alpha + \beta = -(-9)/3 = 3. Product of roots: αβ=2/3\alpha\beta = 2/3. We want to find α2+β2\alpha^2 + \beta^2. We can express this in terms of the sum and product of roots: α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta Substitute the values we found: α2+β2=(3)2−2(2/3)=9−4/3\alpha^2 + \beta^2 = (3)^2 - 2(2/3) = 9 - 4/3 α2+β2=273−43=233\alpha^2 + \beta^2 = \frac{27}{3} - \frac{4}{3} = \frac{23}{3}. Distractor D results from a sign error: (α+β)2+2αβ=9+4/3=31/3(\alpha + \beta)^2 + 2\alpha\beta = 9 + 4/3 = 31/3. Distractor C comes from mistakenly thinking α2+β2=(α+β)2\alpha^2 + \beta^2 = (\alpha+\beta)^2.

Question 13

The graphs of y=x2−2x+5y = x^2 - 2x + 5 and y=−x2+6x+cy = -x^2 + 6x + c intersect at exactly one point. Find the value of cc.

  1. -3 (correct answer)
  2. 3
  3. 5
  4. 13
Explanation: To find the point of intersection, we set the two equations equal to each other: x2−2x+5=−x2+6x+cx^2 - 2x + 5 = -x^2 + 6x + c Rearrange the equation into the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0: 2x2−8x+(5−c)=02x^2 - 8x + (5-c) = 0 For the graphs to intersect at exactly one point, this quadratic equation must have exactly one real solution (a repeated root). This occurs when the discriminant Δ=b2−4ac\Delta = b^2 - 4ac is equal to 0. Here, a=2,b=−8,a=2, b=-8, and the constant term is (5−c)(5-c). Δ=(−8)2−4(2)(5−c)=0\Delta = (-8)^2 - 4(2)(5-c) = 0 64−8(5−c)=064 - 8(5-c) = 0 64−40+8c=064 - 40 + 8c = 0 24+8c=024 + 8c = 0 8c=−248c = -24 c=−3c = -3 Distractor B comes from a sign error when rearranging: 2x2−8x+(5+c)=02x^2-8x+(5+c)=0. Distractor D comes from incorrectly calculating the discriminant as b2+4ac=0b^2+4ac=0.

Question 14

The axis of symmetry of the quadratic function f(x)=2x2+bx−7f(x) = 2x^2 + bx - 7 is x=3x=3. Find the value of bb.

  1. -12 (correct answer)
  2. -6
  3. 6
  4. 12
Explanation: The formula for the axis of symmetry of a quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c is x=−b2ax = -\frac{b}{2a}. For the function f(x)=2x2+bx−7f(x) = 2x^2 + bx - 7, we have a=2a=2. We are given that the axis of symmetry is x=3x=3. So, we can set up the equation: 3=−b2(2)3 = -\frac{b}{2(2)} 3=−b43 = -\frac{b}{4} Multiply both sides by 4: 12=−b12 = -b Therefore, b=−12b = -12. Distractor D results from a sign error. Distractor B results from forgetting the coefficient aa in the denominator of the formula (using x=−b/2x = -b/2).

Question 15

The quadratic expression x2+(k−3)x+kx^2 + (k-3)x + k is positive for all real values of xx. Find the set of possible values for kk.

  1. k<1k < 1 or k>9k > 9
  2. 1<k<91 < k < 9 (correct answer)
  3. k>3k > 3
  4. k=1k=1 or k=9k=9
Explanation: For a quadratic expression ax2+bx+cax^2+bx+c to be always positive, two conditions must be met:
  1. The parabola must open upwards, i.e., a>0a > 0.
  2. The parabola must not intersect the x-axis, i.e., it must have no real roots, which means the discriminant Δ=b2−4ac\Delta = b^2 - 4ac must be negative.
For the expression x2+(k−3)x+kx^2 + (k-3)x + k, we have a=1,b=k−3,c=ka=1, b=k-3, c=k.
  1. Condition 1 is satisfied since a=1>0a=1 > 0.
  2. For condition 2, we need Δ<0\Delta < 0. Δ=(k−3)2−4(1)(k)<0\Delta = (k-3)^2 - 4(1)(k) < 0 k2−6k+9−4k<0k^2 - 6k + 9 - 4k < 0 k2−10k+9<0k^2 - 10k + 9 < 0 Factoring the quadratic in kk: (k−1)(k−9)<0(k-1)(k-9) < 0 The roots are k=1k=1 and k=9k=9. Since this is an upward-opening parabola in kk, the expression is negative between the roots. Therefore, 1<k<91 < k < 9.
Distractor A is the solution for Δ>0\Delta > 0. Distractor D is the solution for Δ=0\Delta = 0.

Question 16

For what value of kk is the line y=2x−1y = 2x - 1 tangent to the parabola y=x2−3x+ky = x^2 - 3x + k?

  1. k=−29/4k = -29/4
  2. k=21/4k = 21/4 (correct answer)
  3. k=25/4k = 25/4
  4. k=29/4k = 29/4
Explanation: If the line is tangent to the parabola, they intersect at exactly one point. We can find the intersection points by setting the equations equal to each other: x2−3x+k=2x−1x^2 - 3x + k = 2x - 1 Rearrange this into a standard quadratic form ax2+bx+c=0ax^2+bx+c=0: x2−5x+(k+1)=0x^2 - 5x + (k+1) = 0 For exactly one point of intersection, this quadratic equation must have exactly one solution (a repeated root). This occurs when the discriminant Δ=b2−4ac\Delta = b^2 - 4ac is equal to zero. Here, a=1,b=−5,c=k+1a=1, b=-5, c=k+1. Δ=(−5)2−4(1)(k+1)=0\Delta = (-5)^2 - 4(1)(k+1) = 0 25−4(k+1)=025 - 4(k+1) = 0 25−4k−4=025 - 4k - 4 = 0 21−4k=021 - 4k = 0 4k=214k = 21 k=214k = \frac{21}{4} Distractor D results from an expansion error: 25−4(k+1)=25−4k+425 - 4(k+1) = 25 - 4k + 4. Distractor A results from using b2+4ac=0b^2+4ac=0.

Question 17

The function f(x)=5−6x−3x2f(x) = 5 - 6x - 3x^2 has a maximum value MM. What is the range of the function?

  1. y≤2y \le 2
  2. y≤8y \le 8 (correct answer)
  3. y≥8y \ge 8
  4. y≥−1y \ge -1
Explanation: The function f(x)=−3x2−6x+5f(x) = -3x^2 - 6x + 5 is a quadratic with a negative leading coefficient (a=−3a=-3), so its graph is a parabola opening downwards. This means the function has a maximum value at its vertex. The x-coordinate of the vertex is given by x=−b2ax = -\frac{b}{2a}. x=−−62(−3)=−−6−6=−1x = -\frac{-6}{2(-3)} = -\frac{-6}{-6} = -1. The maximum value MM is the y-coordinate of the vertex, which is f(−1)f(-1). M=f(−1)=5−6(−1)−3(−1)2=5+6−3=8M = f(-1) = 5 - 6(-1) - 3(-1)^2 = 5 + 6 - 3 = 8. Since the maximum value of the function is 8, the range is all values less than or equal to 8. Therefore, the range is y≤8y \le 8.

Question 18

For what value of kk is the line y=2x−1y = 2x - 1 tangent to the parabola y=x2−3x+ky = x^2 - 3x + k?

  1. k=−29/4k = -29/4
  2. k=21/4k = 21/4 (correct answer)
  3. k=25/4k = 25/4
  4. k=29/4k = 29/4
Explanation: If the line is tangent to the parabola, they intersect at exactly one point. We can find the intersection points by setting the equations equal to each other: x2−3x+k=2x−1x^2 - 3x + k = 2x - 1 Rearrange this into a standard quadratic form ax2+bx+c=0ax^2+bx+c=0: x2−5x+(k+1)=0x^2 - 5x + (k+1) = 0 For exactly one point of intersection, this quadratic equation must have exactly one solution (a repeated root). This occurs when the discriminant Δ=b2−4ac\Delta = b^2 - 4ac is equal to zero. Here, a=1,b=−5,c=k+1a=1, b=-5, c=k+1. Δ=(−5)2−4(1)(k+1)=0\Delta = (-5)^2 - 4(1)(k+1) = 0 25−4(k+1)=025 - 4(k+1) = 0 25−4k−4=025 - 4k - 4 = 0 21−4k=021 - 4k = 0 4k=214k = 21 k=214k = \frac{21}{4} Distractor D results from an expansion error: 25−4(k+1)=25−4k+425 - 4(k+1) = 25 - 4k + 4. Distractor A results from using b2+4ac=0b^2+4ac=0.

Question 19

Find the set of values of xx for which 3x+10<x23x + 10 < x^2.

  1. −2<x<5-2 < x < 5
  2. x<−5x < -5 or x>2x > 2
  3. x<−2x < -2 or x>5x > 5 (correct answer)
  4. x>5x > 5
Explanation: First, rearrange the inequality to the standard quadratic form: x2−3x−10>0x^2 - 3x - 10 > 0 Next, find the roots of the corresponding quadratic equation x2−3x−10=0x^2 - 3x - 10 = 0 by factoring: (x−5)(x+2)=0(x-5)(x+2) = 0 The roots are x=5x=5 and x=−2x=-2. These roots are the critical values where the expression equals zero. We need to find where the expression is greater than zero. The graph of y=x2−3x−10y = x^2 - 3x - 10 is an upward-opening parabola. It will be above the x-axis (i.e., positive) when xx is outside the roots. Therefore, the solution is x<−2x < -2 or x>5x > 5. Distractor A is the solution to x2−3x−10<0x^2 - 3x - 10 < 0. Distractor B results from incorrectly factoring the quadratic as (x+5)(x−2)(x+5)(x-2).

Question 20

The quadratic equation 3x2−9x+2=03x^2 - 9x + 2 = 0 has roots α\alpha and β\beta. Find the value of α2+β2\alpha^2 + \beta^2.

  1. 5/35/3
  2. 23/323/3 (correct answer)
  3. 99
  4. 31/331/3
Explanation: For the quadratic equation ax2+bx+c=0ax^2+bx+c=0, the sum of roots is α+β=−b/a\alpha + \beta = -b/a and the product of roots is αβ=c/a\alpha\beta = c/a. For 3x2−9x+2=03x^2 - 9x + 2 = 0, we have a=3,b=−9,c=2a=3, b=-9, c=2. Sum of roots: α+β=−(−9)/3=3\alpha + \beta = -(-9)/3 = 3. Product of roots: αβ=2/3\alpha\beta = 2/3. We want to find α2+β2\alpha^2 + \beta^2. We can express this in terms of the sum and product of roots: α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta Substitute the values we found: α2+β2=(3)2−2(2/3)=9−4/3\alpha^2 + \beta^2 = (3)^2 - 2(2/3) = 9 - 4/3 α2+β2=273−43=233\alpha^2 + \beta^2 = \frac{27}{3} - \frac{4}{3} = \frac{23}{3}. Distractor D results from a sign error: (α+β)2+2αβ=9+4/3=31/3(\alpha + \beta)^2 + 2\alpha\beta = 9 + 4/3 = 31/3. Distractor C comes from mistakenly thinking α2+β2=(α+β)2\alpha^2 + \beta^2 = (\alpha+\beta)^2.