Question

1. find the center (h, k) and radius r of the circle. graph the circle.\

$x^2 + y^2 - 10x - 2y + 17 = 0$\
graph of coordinate plane\

1. determine whether the graph is that of a function. if it is, use the graph to find its domain and range, the intercepts, if any, and any symmetry with respect to the x-axis, the y-axis, or the origin.\

graph of a curve with x-axis labeled with -π, -3π/4, -π/2, -π/4, 0, π/4, π/2, 3π/4, π\

1. determine algebraically whether the function is even, odd, or neither.\

$f(x) = -5x^2 - 4$

Explanation:

Question 8

Step1: Rewrite the equation

Given the circle equation \(x^{2}+y^{2}-10x - 2y+17 = 0\). Group the \(x\)-terms and \(y\)-terms: \((x^{2}-10x)+(y^{2}-2y)=- 17\).

Step2: Complete the square for \(x\) and \(y\)

For the \(x\)-terms: \(x^{2}-10x=(x - 5)^{2}-25\) (since \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(2ab = 10x\), so \(b = 5\)).
For the \(y\)-terms: \(y^{2}-2y=(y - 1)^{2}-1\) (here \(a = y\), \(2ab=2y\), so \(b = 1\)).

Substitute these back into the equation: \((x - 5)^{2}-25+(y - 1)^{2}-1=-17\).

Step3: Simplify the equation

Simplify the left - hand side: \((x - 5)^{2}+(y - 1)^{2}-26=-17\).
Then, \((x - 5)^{2}+(y - 1)^{2}=26 - 17=9\).

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Comparing \((x - 5)^{2}+(y - 1)^{2}=9=3^{2}\) with the standard form, we get \(h = 5\), \(k = 1\) and \(r = 3\).

Question 9

Step1: Check if it is a function

We use the vertical line test. For any vertical line \(x = a\) (where \(a\in[-\pi,\pi]\)), the vertical line intersects the graph at most once. So the graph represents a function.

Step2: Find the domain

The domain is the set of all \(x\) - values for which the function is defined. From the graph, the left - most point has \(x=-\pi\) and the right - most point has \(x = \pi\). So the domain is \([-\pi,\pi]\).

Step3: Find the range

The range is the set of all \(y\) - values the function takes. The lowest point on the graph has \(y=-1\) and the highest point has \(y = 1\). So the range is \([-1,1]\).

Step4: Find the intercepts

• \(x\) - intercepts: The points where \(y = 0\). From the graph, when \(y = 0\), \(x=-\pi,0,\pi\).
• \(y\) - intercept: The point where \(x = 0\). From the graph, when \(x = 0\), \(y = 0\).

Step5: Check for symmetry

• Symmetry about the \(x\) - axis: If \((x,y)\) is on the graph, then \((x, - y)\) should also be on the graph. If we take a point \((x,y)\) in the upper half (\(y>0\)), the point \((x, - y)\) is also on the graph, but we also check for other symmetries.
• Symmetry about the \(y\) - axis: If \((x,y)\) is on the graph, then \((-x,y)\) should also be on the graph. If we take \(x=\frac{\pi}{2}\), the point \((\frac{\pi}{2},y)\) and \((-\frac{\pi}{2},-y)\) (not \((-\frac{\pi}{2},y)\)) are on the graph, so it is not symmetric about the \(y\) - axis.
• Symmetry about the origin: If \((x,y)\) is on the graph, then \((-x,-y)\) should also be on the graph. For any point \((x,y)\) on the graph, the point \((-x,-y)\) is also on the graph. For example, if \((\frac{\pi}{2},1)\) were on the graph (it's not, but for a general point), the point \((-\frac{\pi}{2},-1)\) is on the graph. So the graph is symmetric about the origin.

Question 10

Step1: Recall the definitions

A function \(y = f(x)\) is even if \(f(-x)=f(x)\) for all \(x\) in the domain of \(f\), and odd if \(f(-x)=-f(x)\) for all \(x\) in the domain of \(f\). The domain of \(f(x)=-5x^{2}-4\) is all real numbers \((-\infty,\infty)\), which is symmetric about the origin.

Step2: Calculate \(f(-x)\)

Substitute \(-x\) into the function: \(f(-x)=-5(-x)^{2}-4\).
Since \((-x)^{2}=x^{2}\), we have \(f(-x)=-5x^{2}-4\).

Step3: Compare \(f(-x)\) with \(f(x)\)

We know that \(f(x)=-5x^{2}-4\). Since \(f(-x)=-5x^{2}-4=f(x)\) for all \(x\in(-\infty,\infty)\), the function \(f(x)\) is even.

Answer:

Center: \((5, 1)\), Radius: \(3\)