Quadratic Functions - IB Mathematics Analysis and Approaches

1. The quadratic function $f(x)=ax^2+bx+c$ has a negative y-intercept and two distinct, positive x-intercepts. Which of the following statements must be true?
   Answer: $a < 0$ and $b > 0$ When analyzing quadratic functions with given intercept properties, you need to visualize the parabola's shape and position to determine the signs of the coefficients. Given that $f(x) = ax^2 + bx + c$ has a negative y-intercept and two distinct, positive x-intercepts, let's think about what this parabola looks like. The y-intercept occurs at $(0, c)$, so $c < 0$. The parabola crosses the x-axis at two positive values, meaning it starts below the x-axis (since $c < 0$), rises to cross the positive x-axis twice, then continues in its parabolic path. For a parabola to start below the x-axis and cross it twice on the positive side, it must open downward. This means $a < 0$. Additionally, since both x-intercepts are positive, the vertex of this downward-opening parabola must also have a positive x-coordinate. The vertex occurs at $x = -\frac{b}{2a}$. Since $a < 0$, for this to be positive, we need $b > 0$. Looking at the options: Choice A suggests $a > 0$ and $c > 0$, but we established $a < 0$ and $c < 0$. Choice B claims $ac > 0$, but since both $a$ and $c$ are negative, $ac > 0$ is actually true, though $b > 0$ is correct - this makes B partially right but incomplete about the signs. Choice C correctly identifies $c < 0$ but wrongly claims $b < 0$. Choice D correctly states $a < 0$ and $b > 0$. Remember: when a quadratic has intercepts in specific regions, sketch the parabola first to visualize the required orientation and vertex position.
2. Find the sum of the solutions to the equation $4^x - 5 \cdot 2^x + 4 = 0$.
   Answer: 2 The equation involves exponential terms. We can rewrite $4^x$ as $(2^2)^x = (2^x)^2$. The equation becomes: $(2^x)^2 - 5(2^x) + 4 = 0$ This is a quadratic equation in terms of $2^x$. Let $u = 2^x$. The equation transforms to: $u^2 - 5u + 4 = 0$ We can solve this quadratic for $u$ by factoring: $(u-1)(u-4) = 0$ This gives two possible values for $u$: $u=1$ or $u=4$. Now, we substitute back $u = 2^x$ to find the values of $x$: Case 1: $2^x = 1$. This implies $x=0$. Case 2: $2^x = 4$. This implies $x=2$. The solutions to the original equation are $x=0$ and $x=2$. The sum of the solutions is $0 + 2 = 2$. Distractor D is the sum of the roots of the quadratic in $u$, not $x$. Distractors A and C are the individual solutions for $x$.
3. The vertex of the parabola $y = x^2 + 4x + c$ is located on the x-axis. What is the value of $c$?
   Answer: 4 If the vertex of a parabola is on the x-axis, it means the parabola touches the x-axis at exactly one point. This corresponds to the quadratic equation having exactly one real root (a repeated root). This occurs when the discriminant $\Delta = b^2 - 4ac$ is equal to zero. For the equation $x^2 + 4x + c = 0$, we have $a=1, b=4, c=c$. $\Delta = (4)^2 - 4(1)(c) = 0$ $16 - 4c = 0$ $16 = 4c$ $c = 4$ Alternatively, we can find the coordinates of the vertex. The x-coordinate is $x = -\frac{b}{2a} = -\frac{4}{2(1)} = -2$. Since the vertex is on the x-axis, its y-coordinate is 0. So the vertex is at $(-2, 0)$. Substitute this point into the parabola's equation: $0 = (-2)^2 + 4(-2) + c$ $0 = 4 - 8 + c$ $0 = -4 + c$ $c = 4$ Distractor A comes from a sign error in the second method. Distractor B is the x-coordinate of the vertex.
4. For which values of $m$ does the equation $x^2 + mx + m + 3 = 0$ have no real roots?
   Answer: $-2 < m < 6$ A quadratic equation has no real roots if its discriminant $\Delta = b^2 - 4ac$ is less than zero. For the equation $x^2 + mx + m + 3 = 0$, we have $a=1, b=m, c=m+3$. We set $\Delta < 0$: $\Delta = m^2 - 4(1)(m+3) < 0$ $m^2 - 4m - 12 < 0$ To solve this quadratic inequality, we first find the roots of $m^2 - 4m - 12 = 0$. Factoring the expression: $(m-6)(m+2) = 0$ The roots are $m=6$ and $m=-2$. The graph of $y = m^2 - 4m - 12$ is an upward-opening parabola. The expression is negative between the roots. Therefore, the solution is $-2 < m < 6$. Distractor A is the solution for $\Delta > 0$ (two distinct real roots). Distractor B results from incorrectly factoring the quadratic as $(m+6)(m-2)$.
5. The roots of the quadratic equation $2x^2 + bx + c = 0$ are $\alpha$ and $\beta$. Given that $\alpha + \beta = 5$ and $\alpha\beta = -3$, find the values of $b$ and $c$.
   Answer: $b = -10, c = -6$ For a quadratic equation $ax^2+bx+c=0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$. In this case, $a=2$. Sum of roots: $\alpha + \beta = -\frac{b}{2}$. We are given $\alpha + \beta = 5$, so $5 = -\frac{b}{2}$, which implies $b = -10$. Product of roots: $\alpha\beta = \frac{c}{2}$. We are given $\alpha\beta = -3$, so $-3 = \frac{c}{2}$, which implies $c = -6$. Thus, $b=-10$ and $c=-6$. Distractor B arises from ignoring the leading coefficient $a=2$. Distractor C has a sign error for $b$. Distractor D has sign errors and ignores the leading coefficient.
6. Find the set of values of $x$ for which $3x + 10 < x^2$.
   Answer: $x < -2$ or $x > 5$ First, rearrange the inequality to the standard quadratic form: $x^2 - 3x - 10 > 0$ Next, find the roots of the corresponding quadratic equation $x^2 - 3x - 10 = 0$ by factoring: $(x-5)(x+2) = 0$ The roots are $x=5$ and $x=-2$. 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 $y = x^2 - 3x - 10$ is an upward-opening parabola. It will be above the x-axis (i.e., positive) when $x$ is outside the roots. Therefore, the solution is $x < -2$ or $x > 5$. Distractor A is the solution to $x^2 - 3x - 10 < 0$. Distractor B results from incorrectly factoring the quadratic as $(x+5)(x-2)$.
7. The graph of a quadratic function $f(x)$ has its vertex at $(3, -8)$ and passes through the point $(5, 4)$. Find the y-intercept of the graph.
   Answer: $19$ Start with the vertex form of a quadratic function: $f(x) = a(x-h)^2 + k$. Given the vertex $(h, k) = (3, -8)$, the equation is $f(x) = a(x-3)^2 - 8$. To find the value of $a$, use the fact that the graph passes through the point $(5, 4)$. Substitute $x=5$ and $f(x)=4$ into the equation: $4 = a(5-3)^2 - 8$ $4 = a(2)^2 - 8$ $4 = 4a - 8$ $12 = 4a$ $a = 3$ So the function is $f(x) = 3(x-3)^2 - 8$. The y-intercept is the value of $f(x)$ when $x=0$. $f(0) = 3(0-3)^2 - 8$ $f(0) = 3(-3)^2 - 8$ $f(0) = 3(9) - 8$ $f(0) = 27 - 8 = 19$. The y-intercept is 19. Distractor A results from failing to square -3: $3(-3)-8 = -17$. Distractor B results from assuming $a=1$: $(0-3)^2-8 = 1$.
8. A quadratic function is given by $f(x) = -3(x-2)^2 + 5$. When written in the form $f(x) = ax^2 + bx + c$, what is the value of $b$?
   Answer: 12 To find the value of $b$, we need to expand the vertex form of the quadratic function into the standard form $ax^2 + bx + c$. Given $f(x) = -3(x-2)^2 + 5$. First, expand the squared term: $(x-2)^2 = x^2 - 4x + 4$. Now substitute this back into the function's equation: $f(x) = -3(x^2 - 4x + 4) + 5$ Distribute the -3: $f(x) = -3x^2 + 12x - 12 + 5$ Combine the constant terms: $f(x) = -3x^2 + 12x - 7$ Comparing this to $f(x) = ax^2 + bx + c$, we can see that $a=-3$, $b=12$, and $c=-7$. The value of $b$ is 12. Distractor A results from a sign error when distributing -3 to the -4x term. Distractor B is the x-coefficient from $(x-2)^2$ before multiplying by $a$.
9. The graphs of $y = x^2 - 2x + 5$ and $y = -x^2 + 6x + c$ intersect at exactly one point. Find the value of $c$.
   Answer: -3 To find the point of intersection, we set the two equations equal to each other: $x^2 - 2x + 5 = -x^2 + 6x + c$ Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$: $2x^2 - 8x + (5-c) = 0$ 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 $\Delta = b^2 - 4ac$ is equal to 0. Here, $a=2, b=-8,$ and the constant term is $(5-c)$. $\Delta = (-8)^2 - 4(2)(5-c) = 0$ $64 - 8(5-c) = 0$ $64 - 40 + 8c = 0$ $24 + 8c = 0$ $8c = -24$ $c = -3$ Distractor B comes from a sign error when rearranging: $2x^2-8x+(5+c)=0$. Distractor D comes from incorrectly calculating the discriminant as $b^2+4ac=0$.
10. The function $f(x) = 2x^2 - 8x + 11$ can be written in the vertex form $f(x) = a(x-h)^2 + k$. What is the minimum value of this function?
    Answer: 3 The minimum value of a parabola that opens upwards (since $a=2>0$) is the y-coordinate of its vertex, $k$. We can find the vertex by completing the square or using the formula $x = -b/(2a)$. Method 1: Completing the square. $f(x) = 2(x^2 - 4x) + 11$ $f(x) = 2(x^2 - 4x + 4 - 4) + 11$ $f(x) = 2((x-2)^2 - 4) + 11$ $f(x) = 2(x-2)^2 - 8 + 11$ $f(x) = 2(x-2)^2 + 3$ The vertex is at $(2, 3)$. The minimum value is 3. Method 2: Using the vertex formula. The x-coordinate of the vertex is $h = -\frac{b}{2a} = -\frac{-8}{2(2)} = \frac{8}{4} = 2$. The minimum value is the y-coordinate, $k = f(h) = f(2)$. $f(2) = 2(2)^2 - 8(2) + 11 = 2(4) - 16 + 11 = 8 - 16 + 11 = 3$. 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).
11. The axis of symmetry of the quadratic function $f(x) = 2x^2 + bx - 7$ is $x=3$. Find the value of $b$.
    Answer: -12 The formula for the axis of symmetry of a quadratic function $f(x) = ax^2 + bx + c$ is $x = -\frac{b}{2a}$. For the function $f(x) = 2x^2 + bx - 7$, we have $a=2$. We are given that the axis of symmetry is $x=3$. So, we can set up the equation: $3 = -\frac{b}{2(2)}$ $3 = -\frac{b}{4}$ Multiply both sides by 4: $12 = -b$ Therefore, $b = -12$. Distractor D results from a sign error. Distractor B results from forgetting the coefficient $a$ in the denominator of the formula (using $x = -b/2$).
12. The equation $kx^2 - 4x + (k-3) = 0$ has two distinct real roots. Which of the following describes the set of possible values for $k$?
    Answer: $-1 < k < 4, k \neq 0$ For a quadratic equation to have two distinct real roots, the discriminant $\Delta = b^2 - 4ac$ must be positive. Here, $a=k, b=-4, c=k-3$. $\Delta = (-4)^2 - 4(k)(k-3) > 0$ $16 - 4k^2 + 12k > 0$ Dividing by -4 and reversing the inequality sign gives: $k^2 - 3k - 4 < 0$ Factoring the quadratic expression: $(k-4)(k+1) < 0$ The roots of $k^2 - 3k - 4 = 0$ are $k=-1$ and $k=4$. Since the parabola opens upwards, the expression is negative between the roots, so $-1 < k < 4$. Additionally, for the equation to be quadratic, the leading coefficient $k$ cannot be zero. Therefore, the complete set of values is $-1 < k < 4, k \neq 0$.
13. The parabola $y = x^2 - 2kx + k$ lies entirely above the x-axis. Which condition must $k$ satisfy?
    Answer: $0 < k < 1$ For a parabola to lie entirely above the x-axis, two conditions must be met: 1. The parabola must open upwards. The coefficient of $x^2$ is $a=1$, which is positive, so this condition is met. 2. The parabola must not intersect the x-axis, which means the quadratic equation $x^2 - 2kx + k = 0$ has no real roots. This happens when the discriminant $\Delta = b^2 - 4ac$ is negative. Here, $a=1, b=-2k, c=k$. $\Delta = (-2k)^2 - 4(1)(k) < 0$ $4k^2 - 4k < 0$ Divide by 4: $k^2 - k < 0$ Factor the expression: $k(k-1) < 0$ The roots of $k(k-1)=0$ are $k=0$ and $k=1$. The expression $k^2-k$ represents an upward-opening parabola in $k$, so it is negative between its roots. Therefore, $0 < k < 1$. Distractor D is the solution for $\Delta > 0$, which corresponds to the parabola having two distinct x-intercepts.
14. The quadratic expression $x^2 + (k-3)x + k$ is positive for all real values of $x$. Find the set of possible values for $k$.
    Answer: $1 < k < 9$ For a quadratic expression $ax^2+bx+c$ to be always positive, two conditions must be met: 1. The parabola must open upwards, i.e., $a > 0$. 2. The parabola must not intersect the x-axis, i.e., it must have no real roots, which means the discriminant $\Delta = b^2 - 4ac$ must be negative. For the expression $x^2 + (k-3)x + k$, we have $a=1, b=k-3, c=k$. 1. Condition 1 is satisfied since $a=1 > 0$. 2. For condition 2, we need $\Delta < 0$. $\Delta = (k-3)^2 - 4(1)(k) < 0$ $k^2 - 6k + 9 - 4k < 0$ $k^2 - 10k + 9 < 0$ Factoring the quadratic in $k$: $(k-1)(k-9) < 0$ The roots are $k=1$ and $k=9$. Since this is an upward-opening parabola in $k$, the expression is negative between the roots. Therefore, $1 < k < 9$. Distractor A is the solution for $\Delta > 0$. Distractor D is the solution for $\Delta = 0$.
15. The function $f(x) = 5 - 6x - 3x^2$ has a maximum value $M$. What is the range of the function?
    Answer: $y \le 8$ The function $f(x) = -3x^2 - 6x + 5$ is a quadratic with a negative leading coefficient ($a=-3$), 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 $x = -\frac{b}{2a}$. $x = -\frac{-6}{2(-3)} = -\frac{-6}{-6} = -1$. The maximum value $M$ is the y-coordinate of the vertex, which is $f(-1)$. $M = f(-1) = 5 - 6(-1) - 3(-1)^2 = 5 + 6 - 3 = 8$. Since the maximum value of the function is 8, the range is all values less than or equal to 8. Therefore, the range is $y \le 8$.
16. The function $f(x) = x^2 - 6x + 13$ is defined for $x \ge 3$. Find the inverse function $f^{-1}(x)$.
    Answer: $f^{-1}(x) = 3 + \sqrt{x-4}$ 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 $f(x) = x^2 - 6x + 13$. Take half of the coefficient of $x$ (which is $-6$), square it: $(-3)^2 = 9$. This gives us $f(x) = (x-3)^2 + 4$. To find the inverse, set $y = (x-3)^2 + 4$ and solve for $x$. Subtract 4: $y-4 = (x-3)^2$. Take the square root: $\pm\sqrt{y-4} = x-3$. This means $x = 3 \pm \sqrt{y-4}$. Here's the crucial step: since the original function is restricted to $x \geq 3$, you need the positive branch. When $x \geq 3$, we have $x-3 \geq 0$, so $x = 3 + \sqrt{y-4}$. Therefore, $f^{-1}(x) = 3 + \sqrt{x-4}$. Option A gives $3 - \sqrt{x-4}$, which would correspond to the left branch of the parabola ($x < 3$), contradicting our domain restriction. Option B has $-3 + \sqrt{x-4}$, which shifts the vertex incorrectly. Option C has $\sqrt{x+4}$ instead of $\sqrt{x-4}$, 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.
17. The quadratic equation $3x^2 - 9x + 2 = 0$ has roots $\alpha$ and $\beta$. Find the value of $\alpha^2 + \beta^2$.
    Answer: $23/3$ For the quadratic equation $ax^2+bx+c=0$, the sum of roots is $\alpha + \beta = -b/a$ and the product of roots is $\alpha\beta = c/a$. For $3x^2 - 9x + 2 = 0$, we have $a=3, b=-9, c=2$. Sum of roots: $\alpha + \beta = -(-9)/3 = 3$. Product of roots: $\alpha\beta = 2/3$. We want to find $\alpha^2 + \beta^2$. We can express this in terms of the sum and product of roots: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ Substitute the values we found: $\alpha^2 + \beta^2 = (3)^2 - 2(2/3) = 9 - 4/3$ $\alpha^2 + \beta^2 = \frac{27}{3} - \frac{4}{3} = \frac{23}{3}$. Distractor D results from a sign error: $(\alpha + \beta)^2 + 2\alpha\beta = 9 + 4/3 = 31/3$. Distractor C comes from mistakenly thinking $\alpha^2 + \beta^2 = (\alpha+\beta)^2$.
18. For what value of $k$ is the line $y = 2x - 1$ tangent to the parabola $y = x^2 - 3x + k$?
    Answer: $k = 21/4$ 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: $x^2 - 3x + k = 2x - 1$ Rearrange this into a standard quadratic form $ax^2+bx+c=0$: $x^2 - 5x + (k+1) = 0$ For exactly one point of intersection, this quadratic equation must have exactly one solution (a repeated root). This occurs when the discriminant $\Delta = b^2 - 4ac$ is equal to zero. Here, $a=1, b=-5, c=k+1$. $\Delta = (-5)^2 - 4(1)(k+1) = 0$ $25 - 4(k+1) = 0$ $25 - 4k - 4 = 0$ $21 - 4k = 0$ $4k = 21$ $k = \frac{21}{4}$ Distractor D results from an expansion error: $25 - 4(k+1) = 25 - 4k + 4$. Distractor A results from using $b^2+4ac=0$.
19. A rectangular garden is to be enclosed by 40 metres of fencing. Find the maximum possible area of the garden in square metres.
    Answer: 100 m$^2$ Let the length of the rectangle be $l$ and the width be $w$. The perimeter is given by $P = 2l + 2w$. We are given that the perimeter is 40 metres: $2l + 2w = 40 \implies l + w = 20$. We can express $l$ in terms of $w$: $l = 20 - w$. The area of the rectangle is $A = l \times w$. Substituting the expression for $l$: $A(w) = (20 - w)w = 20w - w^2$. This is a quadratic function of $w$. Since the coefficient of the $w^2$ term is negative ($-1$), the graph is a downward-opening parabola, and its vertex represents the maximum area. The w-coordinate of the vertex is $w = -\frac{b}{2a} = -\frac{20}{2(-1)} = 10$. When $w = 10$, the length is $l = 20 - 10 = 10$. The shape that maximizes area is a square. The maximum area is $A = 10 \times 10 = 100$ square metres. Distractor A uses the perimeter value. Distractor D is $20^2$, from the sum of length and width.
20. One root of the quadratic equation $x^2 + kx + 27 = 0$ is three times the other root. Find the sum of the possible values of $k$.
    Answer: 0 Let the roots of the equation be $\alpha$ and $3\alpha$. From Vieta's formulas for a quadratic $ax^2+bx+c=0$: Product of roots: $\alpha \cdot 3\alpha = \frac{c}{a}$ Sum of roots: $\alpha + 3\alpha = -\frac{b}{a}$ In our equation, $a=1, b=k, c=27$. Product: $3\alpha^2 = \frac{27}{1} = 27 \implies \alpha^2 = 9 \implies \alpha = \pm 3$. Sum: $4\alpha = -\frac{k}{1} = -k \implies k = -4\alpha$. We have two possible values for $\alpha$, which lead to two possible values for $k$: Case 1: If $\alpha = 3$, then $k = -4(3) = -12$. Case 2: If $\alpha = -3$, then $k = -4(-3) = 12$. The possible values for $k$ are -12 and 12. The sum of these possible values is $-12 + 12 = 0$. Distractors A and C represent finding only one of the possible values for k. Distractor D is the sum of the absolute values of k.

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