Problem 45
Question
For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph. $$ a(x)=\frac{x^{2}+2 x-3}{x^{2}-1} $$
Step-by-Step Solution
Verified Answer
The horizontal intercepts are \(x = -3\) and \(x = 1\), the vertical intercept is \((0, 3)\), the vertical asymptote is \(x = -1\), and the horizontal asymptote is \(y = 1\).
1Step 1: Find the Horizontal Intercepts
Horizontal intercepts occur where the function equals zero, which means the numerator is zero. We solve for the values of \(x\) that make the numerator zero.\[x^2 + 2x - 3 = 0\]This is a quadratic equation that can be factored as \((x + 3)(x - 1) = 0\). Therefore, the horizontal intercepts are \(x = -3\) and \(x = 1\).
2Step 2: Find the Vertical Intercept
To find the vertical intercept, we need to evaluate the function at \(x = 0\).\[a(0) = \frac{0^2 + 2 \cdot 0 - 3}{0^2 - 1} = \frac{-3}{-1} = 3\]So, the vertical intercept is at the point \((0, 3)\).
3Step 3: Find the Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at those points. Set the denominator equal to zero and solve for \(x\).\[x^2 - 1 = 0\]Factoring, we get \((x - 1)(x + 1) = 0\), which gives \(x = 1\) and \(x = -1\). However, since \(x = 1\) is a horizontal intercept, we must exclude it since the numerator is also zero at \(x = 1\). Therefore, the vertical asymptote is at \(x = -1\).
4Step 4: Determine the Horizontal Asymptote
For rational functions where the degree of the numerator equals the degree of the denominator, the horizontal asymptote is given by the ratio of the leading coefficients. Here, both the numerator and denominator have a degree of 2, so the horizontal asymptote is \(y = \frac{1}{1} = 1\).
Key Concepts
Horizontal InterceptsVertical InterceptsAsymptotesDegree of a Polynomial
Horizontal Intercepts
Horizontal intercepts are where the graph crosses the x-axis. To find them in a rational function like \(a(x) = \frac{x^2 + 2x - 3}{x^2 - 1}\), we set the numerator equal to zero because a fraction is zero when its numerator is zero and its denominator is not zero.
To find these intercepts, solve the equation \(x^2 + 2x - 3 = 0\). You can factor this quadratic equation to get \((x + 3)(x - 1) = 0\). From this factorization, we find two solutions: \(x = -3\) and \(x = 1\).
Thus, the horizontal intercepts are at \((-3, 0)\) and \((1, 0)\).
It’s important to remember that horizontal intercepts are simply the points where the graph touches or crosses the x-axis.
To find these intercepts, solve the equation \(x^2 + 2x - 3 = 0\). You can factor this quadratic equation to get \((x + 3)(x - 1) = 0\). From this factorization, we find two solutions: \(x = -3\) and \(x = 1\).
Thus, the horizontal intercepts are at \((-3, 0)\) and \((1, 0)\).
It’s important to remember that horizontal intercepts are simply the points where the graph touches or crosses the x-axis.
Vertical Intercepts
The vertical intercept is where the graph crosses the y-axis. It corresponds to the function's value when \(x = 0\).
To determine this point for our function \(a(x) = \frac{x^2 + 2x - 3}{x^2 - 1}\), simply substitute \(x = 0\) into the function as follows:
This point marks where the curve intersects the vertical axis of the graph.
To determine this point for our function \(a(x) = \frac{x^2 + 2x - 3}{x^2 - 1}\), simply substitute \(x = 0\) into the function as follows:
- \(a(0) = \frac{0^2 + 2 \cdot 0 - 3}{0^2 - 1} = \frac{-3}{-1} = 3\)
This point marks where the curve intersects the vertical axis of the graph.
Asymptotes
Asymptotes are lines that the graph of a function approaches but never quite reaches.
In rational functions, there are horizontal and vertical asymptotes. Vertical asymptotes occur where the denominator is zero (as long as the numerator isn't zero at those points too). For \(a(x)\), set the denominator equal to zero:
Thus, the only vertical asymptote is \(x = -1\). Horizontal asymptotes are determined based on the degrees of the numerator and the denominator. Since both have a degree of 2, the horizontal asymptote is \(y = \frac{1}{1} = 1\).
This line indicates the value the function will approach as \(x\) goes towards infinity.
In rational functions, there are horizontal and vertical asymptotes. Vertical asymptotes occur where the denominator is zero (as long as the numerator isn't zero at those points too). For \(a(x)\), set the denominator equal to zero:
- \(x^2 - 1 = 0\)
Thus, the only vertical asymptote is \(x = -1\). Horizontal asymptotes are determined based on the degrees of the numerator and the denominator. Since both have a degree of 2, the horizontal asymptote is \(y = \frac{1}{1} = 1\).
This line indicates the value the function will approach as \(x\) goes towards infinity.
Degree of a Polynomial
The degree of a polynomial is the highest power of the variable in the expression.
In the function \(a(x) = \frac{x^2 + 2x - 3}{x^2 - 1}\), both the numerator \(x^2 + 2x - 3\) and the denominator \(x^2 - 1\) are polynomials of degree 2, since the highest exponent of \(x\) in both is 2.
In the function \(a(x) = \frac{x^2 + 2x - 3}{x^2 - 1}\), both the numerator \(x^2 + 2x - 3\) and the denominator \(x^2 - 1\) are polynomials of degree 2, since the highest exponent of \(x\) in both is 2.
- This equality of degrees impacts the horizontal asymptote of a rational function.
- When the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.
- In this case, both leading coefficients are 1, so the horizontal asymptote is \(y = 1\).
Other exercises in this chapter
Problem 45
For the following exercises, use a graph to help determine the domain of the functions. $$f(x)=\sqrt{\frac{9-x^{2}}{x+4}}$$
View solution Problem 45
For the following exercises, find all complex solutions (real and non-real). \(2 x^{3}-3 x^{2}+32 x+17=0\)
View solution Problem 45
For the following exercises, graph the polynomial functions. Note \(x\) -and \(y\) -intercepts, multiplicity, and end behavior. $$ k(x)=(x-3)^{3}(x-2)^{2} $$
View solution Problem 46
For the following exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with \(y\) -coordinate
View solution