Let's go through each question one at a time to determine their respective answers.
First Question:
Determine the domain and range of the quadratic function:
\[ f(x) = -6(x+1)^2 - 9 \]
Domain:
Since it is a quadratic function (a polynomial function), the domain is all real numbers.
Domain: \( \mathbb{R} \) or "the real numbers".
Range:
The function opens downwards (due to the negative coefficient), and the vertex represents the maximum value.
- To find the vertex, use the standard form \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex.
- In this case, \((h, k) = (-1, -9)\).
- The maximum value is \( -9 \). Therefore, the range is \( y \leq -9 \).
Answer for the first question:
Domain: \( \mathbb{R} \) or "the real numbers"; Range: \( y \leq -9 \).
Second Question:
Which quadratic function has a range that can be described as \( y \leq 10.5 \)?
- \( f(x) = -6(x+4)^2 + 10.5 \) - opens downwards and has a maximum of \( 10.5 \).
- \( f(x) = 6(x+4)^2 + 10.5 \) - opens upwards, hence, no upper limit.
- \( f(x) = -6(x-10.5)^2 + 4 \) - opens downwards and has vertex at \( (10.5, 4) \), range is \( y \leq 4 \).
- \( f(x) = 10.5(x-4)^2 - 6 \) - opens upwards, no upper limit.
Only the first option \( f(x) = -6(x+4)^2 + 10.5 \) creates a range of \( y \leq 10.5 \).
Answer for the second question:
\( f(x) = -6(x+4)^2 + 10.5 \)
Third Question:
Graph the equation \( y = -5(x)^2 + 1 \). Which of the following statements about the graph is correct?
The standard form is \( y = a(x - h)^2 + k \). Here:
- \( a = -5 \) (indicates it opens downwards),
- \( h = 0 \) (the x-coordinate of the vertex),
- \( k = 1 \) (the y-coordinate of the vertex).
Therefore, the vertex is at \( (0, 1) \) and the parabola opens downwards.
Answer for the third question:
The parabola will open downwards.
Fourth Question:
Use the image to answer the question.
We need the equation of the upward-opening parabola. Judges according to a vertex form \( y = a(x - h)^2 + k \).
Without the actual image, examining the options:
- \( y = 2(x + 4)^2 + 8 \) - opens upwards.
- \( y = 2(x - 4)^2 + 8 \) - opens upwards.
- \( y = -2(x - 4)^2 + 8 \) - opens downwards.
- \( y = 2(x - 4)^2 - 8 \) - opens upwards.
Any of the first two options could potentially fit an upward-opening parabola depending on vertex location. More context would help narrow this down.
Fifth Question:
Use the image to answer the question.
For a concave down parabola, the vertex form looks like \( y = -a(x-h)^2 + k \). Thus:
- \( y = -14(x + 4)^2 + 2 \) - opens downward.
- \( y = -14(x - 4)^2 + 2 \) - opens downward.
- \( y = -14(x + 2)^2 + 4 \) - opens downward.
- \( y = -14(x - 2)^2 + 4 \) - opens downward.
To select one, we'd analyze based on where the vertex appears located within the graph.
Without seeing the graph, it may be difficult; more context is needed on plotting or the vertex coordinates visually represented.
For the most complete answers:
- Domain: \( \mathbb{R} \), Range: \( y \leq -9 \) for the first question.
- \( f(x) = -6(x+4)^2 + 10.5 \) for the second question.
- The parabola will open downwards.
- Further contextual images required to finalize options for the last parts.
Feel free to ask for any specifics or assistance further clarifying specific answers!
3 answers