Graphing Quadratic Functions - Accuplacer Aaf

The axis of symmetry for a parabola is the line $x = 5$ . One point on the parabola is $(2, 7)$ . Which of the following points must also be on the parabola?
Answer: (8, 7) Parabolas are symmetric about their axis of symmetry. The point $(2, 7)$ has an x-coordinate of 2, which is $5 - 3 = 2$ , or 3 units to the left of the axis of symmetry $x = 5$ . Due to symmetry, there must be another point with the same y-coordinate that is 3 units to the right of the axis of symmetry. Its x-coordinate will be $5 + 3 = 8$ . Therefore, the point $(8, 7)$ must also be on the parabola.
The graph of the quadratic function $f(x) = 2x^2 - 12x + 11$ is a parabola. What is the y-coordinate of the vertex of this parabola?
Answer: -7 The x-coordinate of the vertex of a parabola given by $f(x) = ax^2 + bx + c$ is $x = -\frac{b}{2a}$ . For this function, $a=2$ and $b=-12$ , so the x-coordinate is $x = -\frac{-12}{2(2)} = \frac{12}{4} = 3$ . To find the y-coordinate, substitute this x-value back into the function: $f(3) = 2(3)^2 - 12(3) + 11 = 2(9) - 36 + 11 = 18 - 36 + 11 = -7$ . Therefore, the y-coordinate of the vertex is -7.
A quadratic function has x-intercepts at $x = 2$ and $x = 6$ , and passes through the point $(4, -8)$ . What is the y-coordinate of the vertex?
Answer: $-8$ With x-intercepts at 2 and 6, the function has the form $f(x) = a(x - 2)(x - 6)$ . Using the point $(4, -8)$ : $-8 = a(4 - 2)(4 - 6) = a(2)(-2) = -4a$ , so $a = 2$ . The function is $f(x) = 2(x - 2)(x - 6)$ . The vertex x-coordinate is the midpoint of the intercepts: $x = \frac{2 + 6}{2} = 4$ . The y-coordinate is $f(4) = 2(4 - 2)(4 - 6) = 2(2)(-2) = -8$ . Choice A uses $a = 1$ . Choice B is correct. Choice C uses the wrong sign. Choice D uses $a = -2$ .
For the quadratic function $f(x) = a(x-h)^2 + k$ , it is known that $a>0$ and $k<0$ . Which statement accurately describes the x-intercepts of the graph of $f(x)$ ?
Answer: There are two distinct x-intercepts. The parameter $a$ determines the direction the parabola opens. Since $a > 0$ , the parabola opens upwards. The vertex of the parabola is at the point $(h, k)$ . Since $k < 0$ , the vertex is located below the x-axis. A parabola that opens upwards and has its minimum point (the vertex) below the x-axis must cross the x-axis twice. Therefore, there are two distinct x-intercepts.
For which value of $k$ does the quadratic function $f(x) = 2x^2 - 12x + k$ have its minimum value equal to $-10$ ?
Answer: $k = 8$ When you encounter a quadratic function and need to find when its minimum (or maximum) equals a specific value, you're working with the vertex form and properties of parabolas. Since the coefficient of $x^2$ is positive (2), this parabola opens upward and has a minimum value at its vertex. To find the minimum value of $f(x) = 2x^2 - 12x + k$ , first locate the x-coordinate of the vertex using $x = -\frac{b}{2a}$ . Here, $a = 2$ and $b = -12$ , so $x = -\frac{(-12)}{2(2)} = \frac{12}{4} = 3$ . The minimum value occurs at $x = 3$ , so: $f(3) = 2(3)^2 - 12(3) + k = 18 - 36 + k = k - 18$ . Since we want this minimum to equal $-10$ : $k - 18 = -10$ , which gives us $k = 8$ . Looking at the wrong answers: Choice A ( $k = 18$ ) would give a minimum of $18 - 18 = 0$ , not $-10$ . Choice B ( $k = -10$ ) represents a common trap where students confuse the desired minimum value with the parameter value; this would actually give a minimum of $-10 - 18 = -28$ . Choice D ( $k = 28$ ) would produce a minimum of $28 - 18 = 10$ , which has the right magnitude but wrong sign. Study tip: For vertex problems, always find the x-coordinate first using $x = -\frac{b}{2a}$ , then substitute back into the function. The constant term $k$ shifts the parabola vertically, so work backwards from your desired y-value.
A parabola opens downward and has exactly one x-intercept at $x = 4$ . If the parabola passes through the point $(2, -8)$ , what is the y-intercept?
Answer: $(0, -32)$ Since the parabola has exactly one x-intercept at $x = 4$ , the vertex is at $(4, 0)$ and the function has the form $f(x) = a(x - 4)^2$ where $a < 0$ . Using the point $(2, -8)$ : $-8 = a(2 - 4)^2 = a(4) = 4a$ , so $a = -2$ . The function is $f(x) = -2(x - 4)^2$ . The y-intercept is $f(0) = -2(0 - 4)^2 = -2(16) = -32$ . Choice B uses $a = -4$ . Choice C forgets the negative coefficient. Choice D uses $a = 4$ instead of $a = -2$ .
A parabola has a y-intercept at $(0, -6)$ and its vertex is at $(-2, 2)$ . Which of the following values is one of its x-intercepts?
Answer: -3 Start with the vertex form $y = a(x-h)^2 + k$ . Substitute the vertex $(-2, 2)$ : $y = a(x+2)^2 + 2$ . Use the y-intercept $(0, -6)$ to find $a$ : $-6 = a(0+2)^2 + 2$ , which simplifies to $-6 = 4a + 2$ , so $-8 = 4a$ , and $a = -2$ . The equation is $y = -2(x+2)^2 + 2$ . To find the x-intercepts, set $y=0$ : $0 = -2(x+2)^2 + 2$ . This gives $-2 = -2(x+2)^2$ , so $1 = (x+2)^2$ . Taking the square root gives $\pm 1 = x+2$ . The two solutions are $x = -1 - 2 = -3$ and $x = 1 - 2 = -1$ . One of the x-intercepts is -3.
The range of the quadratic function $f(x) = -3x^2 + 6x + 5$ is $(-\infty, k]$ . What is the value of $k$ ?
Answer: 8 The function represents a downward-opening parabola because the leading coefficient (-3) is negative. The range is determined by the maximum value of the function, which is the y-coordinate of the vertex. First, find the x-coordinate of the vertex: $h = -\frac{b}{2a} = -\frac{6}{2(-3)} = -\frac{6}{-6} = 1$ . Then find the y-coordinate: $k = f(1) = -3(1)^2 + 6(1) + 5 = -3 + 6 + 5 = 8$ . Thus, the maximum value is 8, and the range is $(-\infty, 8]$ . The value of $k$ is 8.
The graph of the quadratic function $y = 2(x+3)^2 - 8$ is translated 4 units to the right and 3 units up. What is the vertex of the new, translated parabola?
Answer: (1, -5) The original function is in vertex form, $y = a(x-h)^2 + k$ , where the vertex is $(h, k)$ . For $y = 2(x+3)^2 - 8$ , which is $y = 2(x - (-3))^2 - 8$ , the vertex is $(-3, -8)$ . Translating 4 units to the right adds 4 to the x-coordinate: $-3 + 4 = 1$ . Translating 3 units up adds 3 to the y-coordinate: $-8 + 3 = -5$ . The new vertex is $(1, -5)$ .
The quadratic function $f(x) = -2x^2 + 8x + k$ has its vertex on the x-axis. What is the y-intercept of this function?
Answer: $(0, -8)$ If the vertex is on the x-axis, then the y-coordinate of the vertex is 0. For $f(x) = -2x^2 + 8x + k$ , the vertex occurs at $x = -\frac{8}{2(-2)} = 2$ . The y-coordinate of the vertex is $f(2) = -2(4) + 8(2) + k = 8 + k$ . Setting this equal to 0: $8 + k = 0$ , so $k = -8$ . The y-intercept is $f(0) = k = -8$ , giving the point $(0, -8)$ . Choice B incorrectly uses $k = 8$ . Choice C uses the x-coordinate of the vertex. Choice D incorrectly calculates $f(2)$ as $-4$ .
The minimum value of a quadratic function $f(x)$ is -5. Its axis of symmetry is $x = 2$ . The graph also passes through the point $(4, 3)$ . What is the y-intercept of the graph?
Answer: 3 The minimum value and axis of symmetry give the vertex of the parabola, which is $(2, -5)$ . Using the vertex form $y = a(x-h)^2 + k$ , we have $y = a(x-2)^2 - 5$ . Use the point $(4, 3)$ to find the value of $a$ : $3 = a(4-2)^2 - 5$ . This simplifies to $3 = a(2)^2 - 5$ , or $3 = 4a - 5$ . Adding 5 to both sides gives $8 = 4a$ , so $a=2$ . The equation is $y = 2(x-2)^2 - 5$ . To find the y-intercept, set $x=0$ : $y = 2(0-2)^2 - 5 = 2(-2)^2 - 5 = 2(4) - 5 = 8 - 5 = 3$ . The y-intercept is 3.
The graph of a parabola is given by the function $f(x) = x^2 - 5x - 14$ . What is the sum of its y-intercept and its positive x-intercept?
Answer: -7 To find the y-intercept, set $x=0$ : $f(0) = 0^2 - 5(0) - 14 = -14$ . To find the x-intercepts, set $f(x)=0$ : $x^2 - 5x - 14 = 0$ . Factoring the quadratic gives $(x-7)(x+2) = 0$ . The x-intercepts are $x=7$ and $x=-2$ . The positive x-intercept is 7. The required sum is the y-intercept plus the positive x-intercept: $-14 + 7 = -7$ .
What is the product of the coordinates of the vertex of the function $f(x) = 2x^2 + 8x - 1$ ?
Answer: 18 First, find the coordinates of the vertex $(h, k)$ . The x-coordinate is $h = -\frac{b}{2a} = -\frac{8}{2(2)} = -\frac{8}{4} = -2$ . The y-coordinate is $k = f(-2) = 2(-2)^2 + 8(-2) - 1 = 2(4) - 16 - 1 = 8 - 16 - 1 = -9$ . The vertex is $(-2, -9)$ . The product of the coordinates is $(-2) \times (-9) = 18$ .
The parabola $y = x^2 + bx + c$ has x-intercepts at $(2,0)$ and $(8,0)$ . What are the coordinates of its vertex?
Answer: (5, -9) The x-coordinate of the vertex is the average of the x-intercepts: $h = \frac{2+8}{2} = 5$ . To find the equation, we can use the sum and product of roots. The sum of the roots is $2+8=10$ , and for $y=x^2+bx+c$ , the sum is $-b$ . So $10=-b$ , which means $b=-10$ . The product of the roots is $2 \times 8=16$ , which is equal to $c$ . The equation is $y = x^2 - 10x + 16$ . Now, find the y-coordinate of the vertex by evaluating the function at $x=5$ : $k = (5)^2 - 10(5) + 16 = 25 - 50 + 16 = -9$ . The vertex is $(5, -9)$ .
What is the distance between the x-intercepts of the function $f(x) = 3(x-4)^2 - 48$ ?
Answer: 8 To find the x-intercepts, set $f(x) = 0$ : $3(x-4)^2 - 48 = 0$ . Add 48 to both sides: $3(x-4)^2 = 48$ . Divide by 3: $(x-4)^2 = 16$ . Take the square root of both sides: $x-4 = \pm 4$ . This gives two equations: $x-4 = 4$ , which solves to $x = 8$ , and $x-4 = -4$ , which solves to $x = 0$ . The two x-intercepts are 8 and 0. The distance between them is $|8 - 0| = 8$ .
The graph of the quadratic function $g(x) = -2x^2 - 20x - 40$ has a maximum value. What is this maximum value?
Answer: 10 The maximum value of a downward-opening parabola is the y-coordinate of its vertex. First, find the x-coordinate of the vertex: $h = -\frac{b}{2a} = -\frac{-20}{2(-2)} = \frac{20}{-4} = -5$ . Then, find the y-coordinate by substituting $h$ into the function: $k = g(-5) = -2(-5)^2 - 20(-5) - 40 = -2(25) + 100 - 40 = -50 + 100 - 40 = 10$ . The maximum value is 10.
A parabola has x-intercepts at $x = -1$ and $x = 5$ , and it passes through the point $(1, -16)$ . What is the y-coordinate of its vertex?
Answer: -18 The x-coordinate of the vertex is the midpoint of the x-intercepts: $h = \frac{-1 + 5}{2} = \frac{4}{2} = 2$ . The equation can be written in factored form as $y = a(x+1)(x-5)$ . Use the point $(1, -16)$ to find $a$ : $-16 = a(1+1)(1-5) = a(2)(-4) = -8a$ , so $a=2$ . The equation is $y = 2(x+1)(x-5)$ . To find the y-coordinate of the vertex, evaluate the function at $x=2$ : $k = 2(2+1)(2-5) = 2(3)(-3) = -18$ .
The vertex of the parabola $y = ax^2 + 16x + 2$ is at $x=-4$ . What is the y-coordinate of the vertex?
Answer: -30 The x-coordinate of the vertex is given by $h = -\frac{b}{2a}$ . We are given $h = -4$ , $b = 16$ . So, $-4 = -\frac{16}{2a}$ , which simplifies to $-4 = -\frac{8}{a}$ . Solving for $a$ gives $-4a = -8$ , so $a = 2$ . The function is $y = 2x^2 + 16x + 2$ . To find the y-coordinate of the vertex, substitute $x = -4$ into the function: $k = 2(-4)^2 + 16(-4) + 2 = 2(16) - 64 + 2 = 32 - 64 + 2 = -30$ .
The parabola given by the equation $y = -2x^2 + bx - 9$ has its axis of symmetry at $x = -3$ . What is the value of $b$ ?
Answer: -12 The formula for the axis of symmetry of a parabola $y = ax^2 + bx + c$ is $x = -\frac{b}{2a}$ . We are given that the axis of symmetry is $x = -3$ , and from the equation we have $a = -2$ . Plugging these values into the formula gives $-3 = -\frac{b}{2(-2)}$ . This simplifies to $-3 = -\frac{b}{-4}$ , which is $-3 = \frac{b}{4}$ . Multiplying both sides by 4 gives $b = -12$ .
The function $g(x) = x^2 - 6x + 5$ is transformed to create $h(x) = -(x^2 - 6x + 5) + 12$ . How do the vertices of these two functions compare?
Answer: The vertex of $h(x)$ is 16 units above the vertex of $g(x)$ For $g(x) = x^2 - 6x + 5$ , completing the square: $g(x) = (x - 3)^2 - 4$ , so the vertex is $(3, -4)$ . For $h(x) = -(x^2 - 6x + 5) + 12 = -g(x) + 12$ . Since $g(3) = -4$ , we have $h(3) = -(-4) + 12 = 4 + 12 = 16$ . The vertex of $h(x)$ is at $(3, 12)$ . The vertical distance is $12 - (-4) = 16$ units. Choice A miscalculates the vertex of $g(x)$ . Choice C only accounts for the reflection. Choice D ignores the reflection.

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