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Chapter 4: Problem 97

Find the distance from the vertex of the parabola \(g(x)=-3 x^{2}+6 x+1\) to the center of the circle \(x^{2}+y^{2}+10 x+8 y+32=0\)

Short Answer

Expert verified
The distance is 10.

Step by step solution

01

Identify the vertex of the parabola

To find the vertex of the parabola given by the equation \(g(x) = -3x^2 + 6x + 1\), use the vertex formula \(x = -\frac{b}{2a}\). In this case, \(a = -3\) and \(b = 6\). Plug these values into the formula to get \[ x = -\frac{6}{2(-3)} = 1 \]. Now, substitute \(x = 1\) back into the equation to find the y-coordinate of the vertex: \[ y = -3(1)^2 + 6(1) + 1 = -3 + 6 + 1 = 4 \]. Therefore, the vertex of the parabola is \((1, 4)\).
02

Identify the center of the circle

The equation of the circle is given by \(x^2 + y^2 + 10x + 8y + 32 = 0\). To find the center, rewrite it in standard form by completing the square. First, group the \(x\) and \(y\) terms: \((x^2 + 10x) + (y^2 + 8y) = -32\). Complete the square for both groups:\(x^2 + 10x\) becomes \((x + 5)^2 - 25\) and \(y^2 + 8y\) becomes \((y + 4)^2 - 16\). Substitute these back into the equation:\((x + 5)^2 - 25 + (y + 4)^2 - 16 = -32\) simplifies to \((x + 5)^2 + (y + 4)^2 = 9\). The center of the circle is \((-5, -4)\).
03

Calculate the distance between the vertex and the center

Use the distance formula to find the distance between the vertex of the parabola \((1, 4)\) and the center of the circle \((-5, -4)\): \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. Substituting \(x_1 = 1\), \(y_1 = 4\), \(x_2 = -5\), and \(y_2 = -4\): \[ d = \sqrt{(-5 - 1)^2 + (-4 - 4)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]. Therefore, the distance is 10.

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Key Concepts

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

vertex of parabola
The vertex of a parabola is a crucial point. It represents the peak of the graph if the parabola opens down, and the lowest point if it opens up. The vertex can be found using the vertex formula for a given quadratic equation in the form \(ax^2 + bx + c\). The formula is \(x = -\frac{b}{2a}\).

For our quadratic equation \(g(x) = -3x^2 + 6x + 1\), \(a = -3\) and \(b = 6\). Plugging these values into the vertex formula, we get \(x = -\frac{6}{2(-3)} = 1\).

Once we have the \(x\)-coordinate, substitute it back into the original equation to get the \(y\)-coordinate. Thus, \(y = -3(1)^2 + 6(1) + 1 = 4\). Therefore, the vertex of the parabola is \((1, 4)\).
center of circle
To find the center of a circle given by a general form equation, we need to rewrite it in standard form through a technique called completing the square. The general form of the circle's equation is \(x^2 + y^2 + 10x + 8y + 32 = 0\).

First group the \(x\) and \(y\) terms together: \((x^2 + 10x) + (y^2 + 8y) = -32\). Now, complete the square for each group. \(x^2 + 10x\) becomes \((x + 5)^2 - 25\) and \(y^2 + 8y\) becomes \((y + 4)^2 - 16\).

Substitute these back into the circle equation: \((x + 5)^2 - 25 + (y + 4)^2 - 16 = -32\). Simplifying, we get \((x + 5)^2 + (y + 4)^2 = 9\), indicating the center of the circle is \((-5, -4)\).
distance formula
The distance formula calculates the distance between two points in a coordinate plane. It is derived from the Pythagorean Theorem and is given by \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

For our problem, we need to find the distance between the vertex of the parabola \((1, 4)\) and the center of the circle \((-5, -4)\). Substituting \(x_1 = 1\), \(y_1 = 4\), \(x_2 = -5\), and \(y_2 = -4\) into the formula, we have: \(d = \sqrt{(-5 - 1)^2 + (-4 - 4)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\).

Therefore, the distance between the vertex of the parabola and the center of the circle is \(10\) units.
completing the square
Completing the square is a method used to convert a quadratic expression into a perfect square trinomial. This technique makes it easier to identify key features of graphs and solve quadratic equations.

For the circle鈥檚 equation given as \(x^2 + y^2 + 10x + 8y + 32 = 0\), we first separate the \(x\) and \(y\) terms: \((x^2 + 10x) + (y^2 + 8y) = -32\).

Next, we complete the square for each grouped terms:
  • For \(x^2 + 10x\), add and subtract \(25\) (i.e., \(10/2 = 5\) and \(5^2 = 25\)), giving us \((x + 5)^2 - 25\).
  • For \(y^2 + 8y\), add and subtract \(16\) (i.e., \(8/2 = 4\) and \(4^2 = 16\)), giving us \((y + 4)^2 - 16\).
Now, substitute back and simplify: \((x + 5)^2 - 25 + (y + 4)^2 - 16 = -32\). This simplifies further to: \((x + 5)^2 + (y + 4)^2 = 9\).

The equation now clearly shows the circle鈥檚 center is \((-5, -4)\) and its radius is \(3\) (as the square root of \(9\) is \(3\)).

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Most popular questions from this chapter

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. $$ \text { Graph } x^{2}-4 x+y^{2}+10 y-7=0 $$

(a) find the vertex and the axis of symmetry of each quadratic function, and determine whether the graph is concave up or concave down. (b) Find the y-intercept and the \(x\) -intercepts, if any. (c) Use parts (a) and (b) to graph the function. (d) Find the domain and the range of the quadratic function. (e) Determine where the quadratic function is increasing and where it is decreasing. (f) Determine where \(f(x)>0\) and where \(f(x)<0\) \(f(x)=x^{2}-2 x-3\)

Challenge Problem Runaway Car Using Hooke's Law, we can show that the work \(W\) done in compressing a spring a distance of \(x\) feet from its at-rest position is \(W=\frac{1}{2} k x^{2}\) where \(k\) is a stiffness constant depending on the spring. It can also be shown that the work done by a body in motion before it comes to rest is given by \(\tilde{W}=\frac{w}{2 g} v^{2},\) where \(w=\) weight of the object (in Ib), \(g=\) acceleration due to gravity \(\left(32.2 \mathrm{ft} / \mathrm{s}^{2}\right),\) and \(v=\) object's velocity \((\mathrm{in} \mathrm{ft} / \mathrm{s})\). A parking garage has a spring shock absorber at the end of a ramp to stop runaway cars. The spring has a stiffness constant \(k=9450 \mathrm{lb} / \mathrm{ft}\) and must be able to stop a \(4000-\mathrm{lb}\) car traveling at \(25 \mathrm{mph}\). What is the least compression required of the spring? Express your answer using feet to the nearest tenth. Source: www.sciforums com

Show that the inequality \((x-4)^{2} \leq 0\) has exactly one solution.

Suppose that \(f(x)=x^{2}+2 x-8\) (a) What is the vertex of \(f ?\) (b) What are the \(x\) -intercepts of the graph of \(f ?\) (c) Solve \(f(x)=-8\) for \(x\). What points are on the graph of \(f ?\) (d) Use the information obtained in parts (a)-(c) to \(\operatorname{graph} f(x)=x^{2}+2 x-8\)
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