Problem 52
Question
Graph each pair of functions. Identify the conic section represented by the graph and write each equation in standard form. \(y=\sqrt{36-x^{2}}\) \(y=-\sqrt{36-x^{2}}\)
Step-by-Step Solution
Verified Answer
The graph represents a circle with its equation in standard form being \( (x-0)^{2} + (y-0)^{2} = 6^{2} \)
1Step 1: Graph the equations
Plot both equations on the same graph. Since both equations are square roots, they will start at the point where the expression under the root is zero, in this case, \(x = -6, x = 6\). Because one equation has a negative square root and one has a positive square root, this means that one half of the conic section is above the x-axis and one half is below.
2Step 2: Identify the conic section
Observe the graph. The figure graphed by the equations should be a circle because unlike parabolas, ellipses, or hyperbolas, circles are symmetric about both axis. The square roots indicate that we have half a circle above the x-axis and half a circle below the x-axis, which form a complete circle.
3Step 3: Write the equation in standard form
Since we know the graph is a circle, we write the equation in the standard form of a circle. The standard form of a circle is \( (x-h)^{2} + (y-k)^{2} = r^{2} \). The equation \( y=\sqrt{36-x^{2}} \) and \( y=-\sqrt{36-x^{2}} \) can be combined and represented in the form of \( x^{2} + y^{2} = 36 \), which can be rewritten as \( (x-0)^{2} + (y-0)^{2} = 6^{2} \). So, the standard form of the given equation is \( (x-0)^{2} + (y-0)^{2} = 6^{2} \) meaning, the circle is centered at the origin (0, 0) with radius 6.
Key Concepts
Graphing FunctionsStandard Form of a CircleCoordinate Geometry
Graphing Functions
Graphing functions is a fundamental skill in mathematics, allowing you to visually assess the behavior of equations and understand their relationships. When graphing, you plot points that satisfy the equation on a coordinate plane. For the functions given, \( y = \sqrt{36-x^{2}} \) and \( y = -\sqrt{36-x^{2}} \), the expressions suggest symmetry and bounded domains. Both functions generate semicircles:
Coordinate points and curves are graphed on the x-y plane, illustrating how equations perform visually, often revealing patterns or symmetries that aren't as obvious from the equation alone.
- The first equation, \( y = \sqrt{36-x^{2}} \), yields the top half of a circle, existing only for \( -6 \leq x \leq 6 \).
- The second equation, \( y = -\sqrt{36-x^{2}} \), provides the bottom half over the same domain.
Coordinate points and curves are graphed on the x-y plane, illustrating how equations perform visually, often revealing patterns or symmetries that aren't as obvious from the equation alone.
Standard Form of a Circle
The standard form of a circle's equation is a concise way to represent its geometric properties. This form is given by:\[(x-h)^{2} + (y-k)^{2} = r^{2}\]Here:
This representation tells you that:
- \((h, k)\) is the center of the circle.
- \(r\) is the radius.
This representation tells you that:
- The circle is centered at \((0,0)\), since both \(x\) and \(y\) terms aren't shifted by any constants \(h\) or \(k\).
- The radius \(r\) is \(6\), since the equation equals \(6^2\).
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves describing geometric shapes numerically using a coordinate plane. It blends algebra and geometry to provide a way to study shapes like circles and their properties using equations.
In the context of circles:
In the context of circles:
- A circle's location and size can be pinpointed with precision when expressed in an equation.
- The circle's equation \(x^2 + y^2 = 36\) reveals that all points \((x, y)\) lie at a constant distance (radius \(r\)) from the origin \((0,0)\).
Other exercises in this chapter
Problem 52
Which point is NOT on the graph of \(4 x^{2}-y^{2}=4 ?\) \(\begin{array}{lllll}{\text { A. }(-2,-2 \sqrt{3})} & {\text { B. }(-1,0)} & {\text { C. }(1,0)} & {\t
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What is the standard form of an equation of a hyperbola? Explain how to rewrite \(25 x^{2}-49 y^{2}=1225\) in standard form.
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Write an equation of a parabola with a vertex at \((1,1)\) directrix \(y=-\frac{1}{2}\)
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Write an equation of an ellipse in standard form with center at the origin and with the given characteristics. height \(29,\) width 53
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