QUESTION IMAGE
Question
drag the labels to the correct locations on the table. not all tiles will be used.
match each attribute of a parabola to the correct quadratic function.
vertex: $(-1,4)$ focus: $left(-1,3\frac{3}{4}\
ight)$ focus: $left(1,3\frac{3}{4}\
ight)$ directrix: $y = 3\frac{7}{8}$
directrix: $y = 4\frac{1}{4}$ vertex: $(-1,-4)$ vertex: $(1,4)$ focus: $left(-1,4\frac{1}{8}\
ight)$ directrix: $y = 3\frac{3}{4}$
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To solve this, we use the vertex form of a parabola \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. The focus and directrix depend on \( a \): for a vertical parabola, the focus is \((h, k + \frac{1}{4a})\) and the directrix is \( y = k - \frac{1}{4a} \).
For \( f(x) = -(x - 1)^2 + 4 \):
- Vertex: The form is \( a(x - h)^2 + k \), so \( h = 1 \), \( k = 4 \). Thus, vertex is \((1, 4)\).
- Focus: \( a = -1 \), so \( \frac{1}{4a} = \frac{1}{4(-1)} = -\frac{1}{4} \). Focus: \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \).
- Directrix: \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \)? Wait, no—wait, \( a = -1 \) (downward opening), so directrix is \( y = k - \frac{1}{4a} \)? Wait, correction: For \( a < 0 \) (downward), the focus is below the vertex, directrix above. Formula: \( \text{focus} = (h, k + \frac{1}{4a}) \), \( \text{directrix}: y = k - \frac{1}{4a} \). Wait, \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \). Thus, focus: \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \), directrix: \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \)? No, wait, no—let’s recalculate. Wait, \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \). So focus is \( (h, k + \frac{1}{4a}) = (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \). Directrix: \( y = k - \frac{1}{4a} = 4 - (-\frac{1}{4}) = 4 + \frac{1}{4} = 4\frac{1}{4} \)? Wait, no—if \( a \) is negative, the parabola opens downward, so the directrix is above the vertex. So \( y = k - \frac{1}{4a} \): since \( \frac{1}{4a} \) is negative (because \( a \) is negative), \( - \frac{1}{4a} \) is positive. So directrix is \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \). Wait, but the labels given include \( \text{directrix}: y = 4\frac{1}{4} \)? No, the labels have \( \text{directrix}: y = 3\frac{7}{8} \)? Wait, maybe I made a mistake. Wait, no, the other function is \( f(x) = 2(x + 1)^2 + 4 \) ( \( a = 2 \), upward opening). Let’s check that.
For \( f(x) = 2(x + 1)^2 + 4 \):
- Vertex: \( h = -1 \), \( k = 4 \), so vertex is \((-1, 4)\).
- Focus: \( a = 2 \), so \( \frac{1}{4a} = \frac{1}{8} \). Focus: \( (-1, 4 + \frac{1}{8}) = (-1, 4\frac{1}{8}) \).
- Directrix: \( y = 4 - \frac{1}{8} = 3\frac{7}{8} \).
Wait, let’s redo both:
Function 1: \( f(x) = -(x - 1)^2 + 4 \)
- Vertex: \( (1, 4) \) (from \( (x - 1) \), so \( h = 1 \), \( k = 4 \)).
- \( a = -1 \) (downward opening).
- Focus: \( (h, k + \frac{1}{4a}) = (1, 4 + \frac{1}{4(-1)}) = (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \).
- Directrix: \( y = k - \frac{1}{4a} = 4 - \frac{1}{4(-1)} = 4 + \frac{1}{4} = 4\frac{1}{4} \). Wait, but the labels have \( \text{directrix}: y = 3\frac{7}{8} \)? No, maybe I mixed up. Wait, the other function:
Function 2: \( f(x) = 2(x + 1)^2 + 4 \)
- Vertex: \( (-1, 4) \) (from \( (x + 1) = (x - (-1)) \), so \( h = -1 \), \( k = 4 \)).
- \( a = 2 \) (upward opening).
- Focus: \( (h, k + \frac{1}{4a}) = (-1, 4 + \frac{1}{4(2)}) = (-1, 4 + \frac{1}{8}) = (-1, 4\frac{1}{8}) \).
- Directrix: \( y = k - \frac{1}{4a} = 4 - \frac{1}{8} = 3\frac{7}{8} \).
Now, let’s match the labels:
For \( f(x) = -(x - 1)^2 + 4 \):
- Vertex: \( (1, 4) \)
- Focus: \( (1, 3\frac{3}{4}) \)
- Directrix: \( y = 4\frac{1}{4} \) (Wait, no—wait, \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \), so focus is \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \), directrix is \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \). But the labels have \( \text{directrix}: y = 3\frac{7}{8} \) for the second function. Wait, maybe I miscalculated \( \frac{1}{4a} \) for \( a = 2 \): \( \frac{1}{4(2)} = \frac{1}{8} \), so focus is \( (-1, 4 + \frac{1}{…
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To solve this, we use the vertex form of a parabola \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. The focus and directrix depend on \( a \): for a vertical parabola, the focus is \((h, k + \frac{1}{4a})\) and the directrix is \( y = k - \frac{1}{4a} \).
For \( f(x) = -(x - 1)^2 + 4 \):
- Vertex: The form is \( a(x - h)^2 + k \), so \( h = 1 \), \( k = 4 \). Thus, vertex is \((1, 4)\).
- Focus: \( a = -1 \), so \( \frac{1}{4a} = \frac{1}{4(-1)} = -\frac{1}{4} \). Focus: \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \).
- Directrix: \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \)? Wait, no—wait, \( a = -1 \) (downward opening), so directrix is \( y = k - \frac{1}{4a} \)? Wait, correction: For \( a < 0 \) (downward), the focus is below the vertex, directrix above. Formula: \( \text{focus} = (h, k + \frac{1}{4a}) \), \( \text{directrix}: y = k - \frac{1}{4a} \). Wait, \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \). Thus, focus: \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \), directrix: \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \)? No, wait, no—let’s recalculate. Wait, \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \). So focus is \( (h, k + \frac{1}{4a}) = (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \). Directrix: \( y = k - \frac{1}{4a} = 4 - (-\frac{1}{4}) = 4 + \frac{1}{4} = 4\frac{1}{4} \)? Wait, no—if \( a \) is negative, the parabola opens downward, so the directrix is above the vertex. So \( y = k - \frac{1}{4a} \): since \( \frac{1}{4a} \) is negative (because \( a \) is negative), \( - \frac{1}{4a} \) is positive. So directrix is \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \). Wait, but the labels given include \( \text{directrix}: y = 4\frac{1}{4} \)? No, the labels have \( \text{directrix}: y = 3\frac{7}{8} \)? Wait, maybe I made a mistake. Wait, no, the other function is \( f(x) = 2(x + 1)^2 + 4 \) ( \( a = 2 \), upward opening). Let’s check that.
For \( f(x) = 2(x + 1)^2 + 4 \):
- Vertex: \( h = -1 \), \( k = 4 \), so vertex is \((-1, 4)\).
- Focus: \( a = 2 \), so \( \frac{1}{4a} = \frac{1}{8} \). Focus: \( (-1, 4 + \frac{1}{8}) = (-1, 4\frac{1}{8}) \).
- Directrix: \( y = 4 - \frac{1}{8} = 3\frac{7}{8} \).
Wait, let’s redo both:
Function 1: \( f(x) = -(x - 1)^2 + 4 \)
- Vertex: \( (1, 4) \) (from \( (x - 1) \), so \( h = 1 \), \( k = 4 \)).
- \( a = -1 \) (downward opening).
- Focus: \( (h, k + \frac{1}{4a}) = (1, 4 + \frac{1}{4(-1)}) = (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \).
- Directrix: \( y = k - \frac{1}{4a} = 4 - \frac{1}{4(-1)} = 4 + \frac{1}{4} = 4\frac{1}{4} \). Wait, but the labels have \( \text{directrix}: y = 3\frac{7}{8} \)? No, maybe I mixed up. Wait, the other function:
Function 2: \( f(x) = 2(x + 1)^2 + 4 \)
- Vertex: \( (-1, 4) \) (from \( (x + 1) = (x - (-1)) \), so \( h = -1 \), \( k = 4 \)).
- \( a = 2 \) (upward opening).
- Focus: \( (h, k + \frac{1}{4a}) = (-1, 4 + \frac{1}{4(2)}) = (-1, 4 + \frac{1}{8}) = (-1, 4\frac{1}{8}) \).
- Directrix: \( y = k - \frac{1}{4a} = 4 - \frac{1}{8} = 3\frac{7}{8} \).
Now, let’s match the labels:
For \( f(x) = -(x - 1)^2 + 4 \):
- Vertex: \( (1, 4) \)
- Focus: \( (1, 3\frac{3}{4}) \)
- Directrix: \( y = 4\frac{1}{4} \) (Wait, no—wait, \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \), so focus is \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \), directrix is \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \). But the labels have \( \text{directrix}: y = 3\frac{7}{8} \) for the second function. Wait, maybe I miscalculated \( \frac{1}{4a} \) for \( a = 2 \): \( \frac{1}{4(2)} = \frac{1}{8} \), so focus is \( (-1, 4 + \frac{1}{8}) = (-1, 4\frac{1}{8}) \), directrix \( y = 4 - \frac{1}{8} = 3\frac{7}{8} \). Yes! That matches the label \( \text{directrix}: y = 3\frac{7}{8} \).
Now, summarize:
\( f(x) = -(x - 1)^2 + 4 \):
- Vertex: \( (1, 4) \)
- Focus: \( (1, 3\frac{3}{4}) \)
- Directrix: \( y = 4\frac{1}{4} \) (Wait, no—wait, the labels have \( \text{directrix}: y = 4\frac{1}{4} \)? Wait, the labels include \( \text{directrix}: y = 4\frac{1}{4} \), \( \text{focus}: (1, 3\frac{3}{4}) \), \( \text{vertex}: (1, 4) \).
\( f(x) = 2(x + 1)^2 + 4 \):
- Vertex: \( (-1, 4) \)
- Focus: \( (-1, 4\frac{1}{8}) \)
- Directrix: \( y = 3\frac{7}{8} \)
Matching Labels:
For \( f(x) = -(x - 1)^2 + 4 \):
- Vertex: \( (1, 4) \)
- Focus: \( (1, 3\frac{3}{4}) \)
- Directrix: \( y = 4\frac{1}{4} \) (Wait, no—wait, the label for directrix \( y = 4\frac{1}{4} \) is present? Wait, the labels are:
Labels:
- vertex: \( (-1, 4) \)
- focus: \( (-1, 3\frac{3}{4}) \)
- focus: \( (1, 3\frac{3}{4}) \)
- directrix: \( y = 3\frac{7}{8} \)
- directrix: \( y = 4\frac{1}{4} \)
- vertex: \( (-1, -4) \)
- vertex: \( (1, 4) \)
- focus: \( (-1, 4\frac{1}{8}) \)
- directrix: \( y = 3\frac{3}{4} \)
So:
\( f(x) = -(x - 1)^2 + 4 \):
- Vertex: \( (1, 4) \) (matches label “vertex: (1, 4)”)
- Focus: \( (1, 3\frac{3}{4}) \) (matches label “focus: (1, 3\frac{3}{4})”)
- Directrix: \( y = 4\frac{1}{4} \) (matches label “directrix: \( y = 4\frac{1}{4} \)”)
\( f(x) = 2(x + 1)^2 + 4 \):
- Vertex: \( (-1, 4) \) (matches label “vertex: (-1, 4)”)
- Focus: \( (-1, 4\frac{1}{8}) \) (matches label “focus: (-1, 4\frac{1}{8})”)
- Directrix: \( y = 3\frac{7}{8} \) (matches label “directrix: \( y = 3\frac{7}{8} \)”)
Wait, but let’s confirm with \( a = 2 \):
For \( f(x) = 2(x + 1)^2 + 4 \), \( a = 2 \), so \( \frac{1}{4a} = \frac{1}{8} \). Thus:
- Focus: \( (-1, 4 + \frac{1}{8}) = (-1, 4\frac{1}{8}) \) (correct).
- Directrix: \( y = 4 - \frac{1}{8} = 3\frac{7}{8} \) (correct).
For \( f(x) = -(x - 1)^2 + 4 \), \( a = -1 \), so \( \frac{1}{4a} = -\frac{1}{4} \):
- Focus: \( (1, 4 - \frac{1}{4}) = (1, 3\frac{3}{4}) \) (correct).
- Directrix: \( y = 4 + \frac{1}{4} = 4\frac{1}{4} \) (correct).
Final Matching:
| \( f(x) = -(x - 1)^2 + 4 \) | \( f(x) = 2(x + 1)^2 + 4 \) |
|---|---|
| focus: \( (1, 3\frac{3}{4}) \) | focus: \( (-1, 4\frac{1}{8}) \) |
| directrix: \( y = 4\frac{1}{4} \) | directrix: \( y = 3\frac{7}{8} \) |
(Note: Other labels like \( \text{vertex: } (-1, -4) \), \( \text{focus: } (-1, 3\frac{3}{4}) \), \( \text{directrix: } y = 3\frac{3}{4} \) are not used, as they don’t match either function.)