Question

1. find the minimum or maximum of the following:

\\4x^2 + 16x - 15\\

a) minimum: 33
b) minimum: -31
c) maximum: 33
d) maximum: -31
e) none of the above

Explanation:

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<reasoning>

Find the axis of symmetry for Question 9

Using the Axis of Symmetry knowledge point
\[

$$\begin{aligned} &f(x) = -x^2 + 2x - 10 \quad (a = -1, \, b = 2)\\ &x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1 \end{aligned}$$

\]

Find the extremum for Question 11

Identify the direction and vertex of the parabola \(g(x) = 4x^2 + 16x - 15\).
Since the leading coefficient \(a = 4 > 0\), the parabola opens upward, meaning it has a minimum value at its vertex.
We find the \(x\)-coordinate of the vertex:
\[
x = -\frac{b}{2a} = -\frac{16}{2(4)} = -2
\]
Now, evaluate the function at \(x = -2\) to find the minimum value:
\[

$$\begin{aligned} g(-2) &= 4(-2)^2 + 16(-2) - 15 \\ &= 4(4) - 32 - 15 \\ &= 16 - 32 - 15 = -31 \end{aligned}$$

\]
Thus, the minimum value is \(-31\).

Identify the vertex and graph for Question 13

Using the Parabola Vertex Form knowledge point
\[

$$\begin{aligned} &f(x) = 2(x - 3)^2 - 1\\ &\text{Vertex } (h, k) = (3, -1)\\ &a = 2 > 0 \quad (\text{opens upward}) \end{aligned}$$

\]
Looking at the options:

• Graph (a) has its vertex in the fourth quadrant at \((3, -1)\).
• Graph (b) has its vertex in the fourth quadrant but shifted further right or down, specifically at \((4, -3)\) or similar. Let's inspect the grid:
• In graph (a), the vertex is located at \(x = 3\), \(y = -1\).
• In graph (b), the vertex is located at \(x = 5\), \(y = -2\).

Therefore, graph (a) represents the function \(f(x) = 2(x - 3)^2 - 1\).
</reasoning>

<answer>

No.	Answer
11	(b) Minimum: -31
13	(a)

</answer>

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Answer:

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<reasoning>

Find the axis of symmetry for Question 9

Using the Axis of Symmetry knowledge point
\[

$$\begin{aligned} &f(x) = -x^2 + 2x - 10 \quad (a = -1, \, b = 2)\\ &x = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1 \end{aligned}$$

\]

Find the extremum for Question 11

Identify the direction and vertex of the parabola \(g(x) = 4x^2 + 16x - 15\).
Since the leading coefficient \(a = 4 > 0\), the parabola opens upward, meaning it has a minimum value at its vertex.
We find the \(x\)-coordinate of the vertex:
\[
x = -\frac{b}{2a} = -\frac{16}{2(4)} = -2
\]
Now, evaluate the function at \(x = -2\) to find the minimum value:
\[

$$\begin{aligned} g(-2) &= 4(-2)^2 + 16(-2) - 15 \\ &= 4(4) - 32 - 15 \\ &= 16 - 32 - 15 = -31 \end{aligned}$$

\]
Thus, the minimum value is \(-31\).

Identify the vertex and graph for Question 13

Using the Parabola Vertex Form knowledge point
\[

$$\begin{aligned} &f(x) = 2(x - 3)^2 - 1\\ &\text{Vertex } (h, k) = (3, -1)\\ &a = 2 > 0 \quad (\text{opens upward}) \end{aligned}$$

\]
Looking at the options:

• Graph (a) has its vertex in the fourth quadrant at \((3, -1)\).
• Graph (b) has its vertex in the fourth quadrant but shifted further right or down, specifically at \((4, -3)\) or similar. Let's inspect the grid:
• In graph (a), the vertex is located at \(x = 3\), \(y = -1\).
• In graph (b), the vertex is located at \(x = 5\), \(y = -2\).

Therefore, graph (a) represents the function \(f(x) = 2(x - 3)^2 - 1\).
</reasoning>

<answer>

No.	Answer
11	(b) Minimum: -31
13	(a)

</answer>

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