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Chapter 1: Problem 24

Do the graphs intersect in the given viewing rectangle? If they do, how many points of intersection are there? $$y=\sqrt{49-x^{2}}, y=\frac{1}{5}(41-3 x) ; \quad[-8,8] \text { by }[-1,8]$$

Short Answer

Expert verified
There are two intersection points within the viewing rectangle.

Step by step solution

01

Understand the Functions

The first function, \( y = \sqrt{49 - x^2} \), is the equation of a circle centered at the origin with a radius of 7, but only the top half of it (since it involves a square root). The second function, \( y = \frac{1}{5}(41 - 3x) \), is a linear equation representing a line with a slope of \(-\frac{3}{5}\) and a y-intercept of \(\frac{41}{5}\).
02

Set Equations Equal to Find Intersections

To find intersection points, we solve for \( x \) where \( \sqrt{49 - x^2} = \frac{1}{5}(41 - 3x) \). First, simplify: \( 5\sqrt{49 - x^2} = 41 - 3x \).
03

Square Both Sides

Square both sides to eliminate the square root: \[ 25(49 - x^2) = (41 - 3x)^2 \]. This leads to \( 1225 - 25x^2 = 1681 - 246x + 9x^2 \).
04

Simplify and Rearrange the Equation

Rearrange and simplify the equation: \(-34x^2 + 246x - 456 = 0\), then divide everything by 2 to simplify: \(-17x^2 + 123x - 228 = 0\).
05

Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -17 \), \( b = 123 \), and \( c = -228 \). Calculate \( \Delta = b^2 - 4ac = 123^2 - 4(-17)(-228) \). Evaluate \( \Delta \) to find the roots of the equation.
06

Calculate the Discriminant

Calculate \( \Delta = 15129 - 4 \times 17 \times 228 \). Compute this to find \( \Delta = 289 \).
07

Find the Roots

Substitute \( \Delta = 289 \) into the quadratic formula: \( x = \frac{-123 \pm 17}{-34} \). This gives two possible \( x \) values. Calculate these values.
08

Determine Intersection Points Within Range

Check if the \( x \) values calculated fall within the interval \([-8, 8]\). If they do, compute corresponding \( y \) values using either equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circle Equation
The equation of a circle is usually written in the form: \[x^2 + y^2 = r^2\]Here, \(r\) represents the radius of the circle. A standard circle centered on the origin \( (0,0) \) is derived from this equation. In the exercise provided, the circle equation is given as \( y = \sqrt{49 - x^2} \). This represents only the upper half of a circle. Why only the upper half? Because by taking the square root, we consider only the positive values of \(y\).
  • The center is at the origin \((0, 0)\).
  • The radius is 7, since \( \sqrt{49} = 7 \).
Visualizing these components is crucial for understanding where potential intersections with other graphs could occur.
Linear Equation
Linear equations represent straight lines. These can be given in several forms: slope-intercept, point-slope, or standard form. For this problem, we're dealing with a slope-intercept form given as:\[y = \frac{1}{5}(41 - 3x)\]This represents a line with:
  • A slope of \(-\frac{3}{5}\), indicating how steep the line climbs or descends.
  • A y-intercept of \frac{41}{5}\, showing where the line crosses the y-axis.
By setting this line equation equal to the equation of the circle, we can find if and where they intersect. This forms the foundation for solving such problems in analytical geometry.
Quadratic Formula
When equations become quadratic, the quadratic formula offers a way to find their roots or solutions. It's expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To solve our equation for intersection, we must transform the expression we derived by setting the circle equation equal to the linear equation, resulting in a quadratic form:
\[-17x^2 + 123x - 228 = 0\]Here:
  • \(a = -17\)
  • \(b = 123\)
  • \(c = -228\)
By substituting these values into the quadratic formula, we can determine potential points of intersection between our two graphs.
Discriminant Calculation
The discriminant, noted as \(\Delta\), plays a big role in determining the nature of the solutions of a quadratic equation. It is found using the formula:\[\Delta = b^2 - 4ac\]In our problem, this calculation shows whether the quadratic equation has real and distinct, real and repeated, or imaginary roots:
  • If \(\Delta > 0\), two distinct real roots exist.
  • If \(\Delta = 0\), exactly one real root is present (a double root).
  • If \(\Delta < 0\), no real roots, implying no intersection within real numbers.
For the circle and line intersection, \(\Delta\) turned out to be 289, which means there are two distinct points of intersection. This significant outcome allows us to know there are exactly two real intersection points.

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Most popular questions from this chapter

(a) Find the radius of each circle in the pair, and the distance between their centers; then use this information to determine whether the circles intersect. (i) \((x-2)^{2}+(y-1)^{2}=9\) \((x-6)^{2}+(y-4)^{2}=16\) (ii) \(x^{2}+(y-2)^{2}=4\) \((x-5)^{2}+(y-14)^{2}=9\) (iii) \((x-3)^{2}+(y+1)^{2}=1\) \((x-2)^{2}+(y-2)^{2}=25\) (b) How can you tell, just by knowing the radii of two circles and the distance between their centers, whether the circles intersect? Write a short paragraph describing how you would decide this and draw graphs to illustrate your answer.

Factor the trinomial. $$2(a+b)^{2}+5(a+b)-3$$

Factor the expression completely. $$x^{6}-8 y^{3}$$

Perform the indicated operations and simplify. $$(1-b)^{2}(1+b)^{2}$$

Factor the expression completely. $$y^{4}(y+2)^{3}+y^{5}(y+2)^{4}$$
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