Question

For each function, predict the locations of any vertical asymptotes, points of discontinuity, and intercepts. Then, graph the function to verify your predictions. a) \(y=\frac{x^{2}+4 x}{x^{2}+9 x+20}\) b) \(y=\frac{2 x^{2}-5 x-3}{x^{2}-1}\) c) \(y=\frac{x^{2}+2 x-8}{x^{2}-2 x-8}\) d) \(y=\frac{2 x^{2}+7 x-15}{9-4 x^{2}}\)

Short answer

a) Vert. Asym.: x=-4, -5. Discont.: x=0, -4. Int.: x=0, -4, y=0.b) Vert. Asym.: x=±1. Discont.: x=-1/2, 3. Int.: x=-1/2, 3, y=3.c) Vert. Asym: x=-4,2 Discont: x^2 + 2 x-8, 0 x intercept: downtown thely y= 1.d) Vert. Asym: check the discriminant: x- y intercept y= -15.

Step-by-step solution

Identify vertical asymptotes

Vertical asymptotes occur where the denominator is zero but the numerator is not zero.

Find points of discontinuity

Points of discontinuity occur where both the numerator and the denominator are zero. These are points where the function is undefined.

Determine intercepts

To find the x-intercepts, set the numerator equal to zero and solve for x. To find the y-intercept, evaluate the function at x=0.

Solve for function a) \(y=\frac{x^{2}+4 x}{x^{2}+9 x+20}\)

1. Factor the denominator: \(x^{2}+9 x+20 = (x+4)(x+5)\). 2. Vertical asymptotes are solutions to \(x^{2}+9 x+20 = 0 \Rightarrow x = -4, -5\). 3. Check for points of discontinuity: Set the numerator equal to zero, \(x(x+4)=0 \Rightarrow x=0, -4\). 4. Check for intercepts: The x-intercepts are at \(x=0, -4 \), and the y-intercept is \(y=0\).

Solve for function b) \(y=\frac{2 x^{2}-5 x-3}{x^{2}-1}\)

1. Factor the denominator: \(x^{2}-1 = (x-1)(x+1)\). 2. Vertical asymptotes are solutions to \(x^{2}-1 = 0 \Rightarrow x = \pm 1)\). 3. Check for points of discontinuity: Set the numerator equal to zero, \(2 x^{2}-5 x-3=0 \Rightarrow (2x+1)(x-3)=0 \ x = -1/2, 3\). 4. Check for intercepts: The x-intercepts are at \(x=-1/2, 3\), and the y-intercept is \(y=3\).

Solve for function c) \(y=\frac{x^{2}+2 x-8}{x^{2}-2 x-8}\)

1. Factor both numerator and denominator: \(x^{2}+2 x-8 = (x+4)(x-2)\) and \(x^{2}-2 x-8 = (x+4)(x-2)\). 2. Vertical asymptotes: \(x^{2}-2 x-8 = 0 \Rightarrow x = -4, 2\)3. Check for points of discontinuity: \Both numerator and denominator are 0. Due to this, every x isn't a point of discontinuity. \4. Intercepts: The x-intercepts are \x = -4, 2 \, y-intercept: \y=1\

Solve for function d) \(y=\frac{2 x^{2}+7 x-15}{9-4 x^{2}}\)

1. Factor the denominator: \9-4 x^{2} = (3-2x)(3+2x)\). 2. Vertical asymptotes: Solve \(3-2x = 0 => 16x - 15 = 0 \. 3.Check for discontinuity: Check solution of \2 x^{2} + 7x - 15 = (2x-3)(x+5)4. Intercepts and, lastly y- intercept: \y= (-3/2, -5) Finally check your graph as cross checking.

Key concepts

Vertical Asymptotes

Vertical asymptotes occur at values of x where the denominator of a rational function is zero, but the numerator is non-zero. This creates points where the function heads towards infinity. To find vertical asymptotes, set the denominator equal to zero and solve for x.

For function a) \(y=\frac{x^{2}+4x}{x^{2}+9x+20}\), first factor the denominator: \(x^{2}+9x+20=(x+4)(x+5)\). Setting \(x+4=0\) and \(x+5=0\) gives vertical asymptotes at \(x=-4\) and \(x=-5\).

For example, in function b) \(y=\frac{2x^{2}-5x-3}{x^{2}-1}\), factor the denominator: \(x^{2}-1=(x-1)(x+1)\), so the vertical asymptotes occur at \(x=\frac{1}{1}, \frac{-1}{1}\). Remember, the numerator must not be zero for these values.

Points of Discontinuity

Points of discontinuity happen where both the numerator and denominator are zero. These points make the function undefined. To identify them, set both the numerator and denominator to zero.

For example, in function c) \(y=\frac{x^{2}+2x-8}{x^{2}-2x-8}\), factor both numerator and denominator: \(x^{2}+2x-8=(x+4)(x-2)\) and \(x^{2}-2x-8=(x+4)(x-2)\). Here, \(x+4\) and \(x-2\) are common factors which make the numerator and denominator zero simultaneously, revealing points of discontinuity at \(x=-4\) and \(x=2\).

In practical terms, these will not show a vertical asymptote but a 'hole' in the graph.

Intercepts

Intercepts tell us where the graph crosses the axes. To find x-intercepts, set the numerator equal to zero and solve for x. To find the y-intercept, evaluate the function at \(x=0\).

Here's a breakdown:
- For function a) \(y=\frac{x^{2}+4x}{x^{2}+9x+20}\), setting the numerator \(x^{2}+4x=0\) gives x-intercepts at \(x=0,-4\). Evaluating at \(x=0\) gives \(y=0\), so the y-intercept is also at \(0\).
- For function b), \(y=\frac{2x^{2}-5x-3}{x^{2}-1}\), setting the numerator \(2x^{2}-5x-3=0\) provides x-intercepts at \(x=-\frac{1}{2}, 3\). Evaluating at \(x=0\) gives \(y=3\).
- Lastly, for function c) \(y=\frac{x^{2}+2x-8}{x^{2}-2x-8}\), setting \(x^{2}+2x-8=0\) provides x-intercepts at \(x=-4,2\). Evaluating at \(x=0\) gives us \(y=1\).

Function Graphing

After computing vertical asymptotes, points of discontinuity, and intercepts, the final task is graphing the function. Graphing these functions helps visualize the behavior of the function near important points.

When graphing rational functions, remember:
- **Vertical Asymptotes:** Draw a dashed vertical line where the asymptote occurs.
- **Intercepts:** Plot points where the graph crosses the x and y axes.
- **Points of Discontinuity:** Indicate holes with open circles.

For example, for function d) \(y=\frac{2x^{2}+7x-15}{9-4x^{2}}\),:
1. Find the vertical asymptotes. Factor \(9-4x^{2}=(3-2x)(3+2x)\), suggesting asymptotes at \(x=\frac{3}{2},-\frac{3}{2}\).
2. Check numerator \(2x^{2}+7x-15=(2x-3)(x+5)\).
3. Finally, compute intercepts, \(x=-\frac{3}{2},-5\).
4. For exact values, apply in the formula.

Visual confirmation of your findings via a graph solidifies understanding and ensures accuracy!