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Chapter 0: Problem 20

Finding Intercepts In Exercises \(17-26,\) find any intercepts. $$ y^{2}=x^{3}-4 x $$

Short Answer

Expert verified
The x-intercepts of the function \(y^{2}=x^{3}-4x\) are \(0, 2, -2\) and the y-intercept is \(0\).

Step by step solution

01

Finding the x-intercepts

Set \(y = 0\) in the equation to find the x-intercepts. So the equation will be\:\[0 = x^{3} - 4x\]Or equivalently\:\[x^{3} - 4x = 0\]This equation can be factored into\:\[x(x^2 - 4) = 0\]or\[x(x - 2)(x + 2) = 0\]Therefore, \(x = 0\), \(x = 2\), and \(x = -2\) are the x-intercepts.
02

Finding the y-intercepts

Set \(x = 0\) into the equation to find the y-intercepts. The resulting equation is\:\[y^2 = 0^3 - 4(0)\]So \(y^2 = 0\), and hence the y-intercept is \(y = 0\).

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Key Concepts

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

Understanding x-intercepts
To find the x-intercepts of a polynomial equation, you need to solve the equation by setting the variable \( y \) to zero. This gives you the points where the graph crosses or touches the x-axis. For the equation \( y^2 = x^3 - 4x \), set \( y = 0 \) which transforms the equation into \( 0 = x^3 - 4x \).

When this equation simplifies to \( x(x^2 - 4) = 0 \), you're looking for all values of \( x \) that make the equation true. This step involves factoring, which we'll cover more deeply in another section. The factored form, \( x(x-2)(x+2) = 0 \), shows us that the x-values of "0," "2," and "-2" make the equation hold true. Hence, these are your x-intercepts.
Grasping y-intercepts
Y-intercepts are found by setting the variable \( x \) to zero in your equation. When you do this, you find where the graph crosses the y-axis. In the provided equation, \( y^2 = x^3 - 4x \), setting \( x = 0 \) makes the equation become \( y^2 = 0 \).

Solving \( y^2 = 0 \) yields \( y = 0 \). Therefore, the y-intercept here is at the origin, or \( (0, 0) \). Note that even though different equations might have multiple y-intercepts, in this case, the single value of y leads us to just one y-intercept.
Demystifying Factoring Equations
Factoring is a crucial skill in solving polynomial equations, helping to find intercepts more easily. Factoring means breaking down an equation into simpler multiplicative components. Once broken down, this makes it straightforward to find where the equation equals zero.

In our example with \( x^3 - 4x = 0 \), common factoring transforms it to \( x(x^2 - 4) = 0 \). Next, further factoring \( x^2 - 4 \) using the difference of squares rule gives us \( (x - 2)(x + 2) \).

This result, \( x(x - 2)(x + 2) = 0 \), makes it clear that if any factor is zero, the whole equation is zero. That's why solutions \( x = 0, x = 2, \) and \( x = -2 \) provide the x-intercepts.
Introduction to Polynomial Equations
Polynomial equations are expressions involving a sum of powers of variables. For instance, an equation like \( y^2 = x^3 - 4x \) is considered a polynomial because it includes powers of x and y.

The degree of the polynomial is the highest power present; in this case, the degree is 3 (from \( x^3 \)). Such equations can have significant features: roots, intercepts, and turning points, which describe how their graphs behave on a coordinate plane.
  • Roots: Values of \( x \) when the polynomial is zero.
  • Intercepts: Points where the graph intersects the axes.
  • Turning Points: Points where the graph changes direction.
Understanding the interplay of these components lays a strong foundation for solving various polynomial problems.

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Most popular questions from this chapter

Finding the Domain and Range of a Function In Exercises \(11-22,\) find the domain and range of the function. $$ g(x)=x^{2}-5 $$

Distance Show that the distance between the point \(\left(x_{1}, y_{1}\right)\) and the line \(A x+B y+C=0\) is Distance \(=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)

Straight-Line Depreciation A small business purchases a piece of equipment for \(\$ 875\) . After 5 years, the equipment will be outdated, having no value. (a) Write a linear equation giving the value \(y\) of the equipment in terms of the time \(x\) (in years), \(0 \leq x \leq 5 .\) (b) Find the value of the equipment when \(x=2\) (c) Estimate (to two-decimal-place accuracy) the time when the value of the equipment is \(\$ 200\) .

Sketching a Graph of a Function In Exercises \(33-40\) , sketch a graph of the function and find its domain and range. Use a graphing utility to verify your graph. $$ g(x)=\frac{4}{x} $$

Distance In Exercises \(83-86\) , find the distance between the point and line, or between the lines, using the formula for the distance between the point \(\left(x_{1}, y_{1}\right)\) and the line \(A x+B y+\) \(C=0 .\) $$ =\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}} $$ Point: \((-2,1)\) Line: \(x-y-2=0\)
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