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Chapter 0: Problem 19

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

Short Answer

Expert verified
The y-intercept is -2 and the x-intercepts are 1 and -2.

Step by step solution

01

Find y-intercept

Set x to be 0 in the function: \( y = 0^{2} + 0 - 2 = -2 \). Thus, the y-intercept is -2.
02

Find x-intercept

Set y to be 0 in the function and solve for x: \( 0 = x^2 + x -2 \). This is a quadratic equation that can be factored into \( 0 = (x - 1)(x + 2) \), which gives two solutions for x, namely x = 1 and x = -2. Thus, the x-intercepts are 1 and -2.

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Key Concepts

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

Quadratic Functions
A quadratic function is a type of polynomial function that is characterized by its highest power being a square. It takes the general form of \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The unique structure of quadratic functions gives them a distinct U-shape graph known as a parabola.
  • The direction of the parabola depends on the sign of \( a \). If \( a > 0 \), the parabola opens upwards, resembling a smile. Conversely, if \( a < 0 \), it opens downwards, like a frown.
  • The vertex of the parabola is its turning point and provides important information about the maximum or minimum value of the function.
Quadratic functions are often explored to determine intercepts, which show where the graph intersects the axes. This helps in analyzing where the output value is zero (x-intercepts) or where the input is zero (y-intercept). Becoming familiar with these characteristics enables us to solve many practical problems involving parabolas.
Quadratic Equations
When dealing with quadratic functions, you frequently encounter quadratic equations. These are equations that set a quadratic expression equal to zero, typically in the form \( ax^2 + bx + c = 0 \). Solving quadratic equations is all about finding the roots, or x-intercepts, of the corresponding quadratic function.
  • Solutions to quadratic equations are often found using various methods such as factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Every quadratic equation has at most two solutions, which could be real or complex numbers, depending on the discriminant \( b^2 - 4ac \).
For example, in the equation \( x^2 + x - 2 = 0 \), you can find the solutions by solving for x through factoring, resulting in two x-intercepts: 1 and -2. Understanding how to manipulate and solve these equations is a key skill in algebra and helps in unearthing insights from the quadratic function's graph.
Factoring Polynomials
Factoring is a fundamental skill in algebra that simplifies the process of solving polynomial equations. For quadratic equations, this technique involves expressing the quadratic expression as a product of two binomials if possible. This aids in finding the roots of the polynomial by setting each factor equal to zero.
  • To factor a quadratic like \( x^2 + x - 2 \), you first look for two numbers that multiply to the constant term (\(-2\)) and add up to the coefficient of the linear term (\(1\)).
  • In this instance, the numbers -1 and 2 fit the bill since \(-1 \times 2 = -2\) and \(-1 + 2 = 1\).
Therefore, \( x^2 + x - 2 \) can be factored into \((x - 1)(x + 2)\). Solving \((x - 1)(x + 2) = 0\) gives the x-intercepts, 1 and -2. Factoring not only simplifies solving quadratic equations but also aids in understanding the graph's behavior and positioning.

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Most popular questions from this chapter

Finding the Domain of a Function In Exercises \(23-28\) , find the domain of the function. $$ f(x)=\sqrt{x}+\sqrt{1-x} $$

Finding the Domain and Range of a Piecewise Function In Exercises \(29-32,\) evaluate the function as indicated. Determine its domain and range. $$ f(x)=\left\\{\begin{array}{ll}{x^{2}+2,} & {x \leq 1} \\ {2 x^{2}+2,} & {x>1}\end{array}\right. $$ $$ \begin{array}{llll}{\text { (a) } f(-2)} & {\text { (b) } f(0)} & {\text { (c) } f(1)} & {\text { (d) } f\left(s^{2}+2\right)}\end{array} $$

Tangent Line Find an equation of the line tangent to the circle \(x^{2}+y^{2}=169\) at the point \((5,12)\) .

Finding the Domain and Range of a Function In Exercises \(11-22,\) find the domain and range of the function. $$ g(x)=\sqrt{6 x} $$

Consider a polynomial \(f(x)\) with real coefficients having the property \(f(g(x))=g(f(x))\) for every polynomial \(g(x)\) with real coefficients. Determine and prove the nature of \(f(x)\)
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