Math Problem Statement

The function f(x) = (6x^2) / (x^2 + 9) is analyzed, with questions involving the point (-√3, 1), domain, intercepts, and calculations for specific values of x.

Solution

Let's analyze the image step by step and solve the questions regarding the function f(x)=6x2x2+9f(x) = \frac{6x^2}{x^2 + 9}.

Questions from the image:

(a) Is the point (3,1)(- \sqrt{3}, 1) on the graph of ff?

We need to substitute x=3x = -\sqrt{3} into the function and check if the result equals 1.

f(3)=6(3)2(3)2+9=6(3)3+9=1812=1f(-\sqrt{3}) = \frac{6(-\sqrt{3})^2}{(-\sqrt{3})^2 + 9} = \frac{6(3)}{3 + 9} = \frac{18}{12} = 1 Yes, the point (3,1)(- \sqrt{3}, 1) is on the graph of ff.

(b) If x=1x = 1, what is f(x)f(x)? What point is on the graph of ff?

Substitute x=1x = 1 into the function: f(1)=6(1)2(1)2+9=6(1)1+9=610=35f(1) = \frac{6(1)^2}{(1)^2 + 9} = \frac{6(1)}{1 + 9} = \frac{6}{10} = \frac{3}{5} So, the point on the graph when x=1x = 1 is (1,35)(1, \frac{3}{5}).

(c) If f(x)=1f(x) = 1, what is xx? What point(s) are on the graph of ff?

Set f(x)=1f(x) = 1 and solve for xx: 6x2x2+9=1\frac{6x^2}{x^2 + 9} = 1 Multiply both sides by x2+9x^2 + 9: 6x2=x2+96x^2 = x^2 + 9 Rearrange the equation: 6x2x2=9    5x2=9    x2=956x^2 - x^2 = 9 \implies 5x^2 = 9 \implies x^2 = \frac{9}{5} Thus: x=±95=±35=±355x = \pm \sqrt{\frac{9}{5}} = \pm \frac{3}{\sqrt{5}} = \pm \frac{3\sqrt{5}}{5} The points on the graph are (355,1)\left(\frac{3\sqrt{5}}{5}, 1\right) and (355,1)\left(-\frac{3\sqrt{5}}{5}, 1\right).

(d) What is the domain of ff?

Since the function is a rational function with no restrictions (no values that make the denominator zero), the domain is all real numbers: Domain of f(x):(,)\text{Domain of } f(x): (-\infty, \infty)

(e) List the x-intercepts, if any, of the graph of ff.

To find the x-intercepts, set f(x)=0f(x) = 0: 6x2x2+9=0\frac{6x^2}{x^2 + 9} = 0 This implies that 6x2=06x^2 = 0, which results in x=0x = 0. Thus, the x-intercept is (0,0)(0, 0).

(f) List the y-intercept, if there is one, of the graph of ff.

The y-intercept occurs when x=0x = 0: f(0)=6(0)2(0)2+9=09=0f(0) = \frac{6(0)^2}{(0)^2 + 9} = \frac{0}{9} = 0 Thus, the y-intercept is (0,0)(0, 0).


Do you need more details on any part of this explanation?

5 Related Questions:

  1. How do we find vertical and horizontal asymptotes of this rational function?
  2. Can you explain how to calculate the derivative of f(x)f(x) to find critical points?
  3. How does the behavior of the function change as xx approaches infinity or negative infinity?
  4. What is the range of the function f(x)f(x)?
  5. How would you graph the function and its key points?

Tip:

For rational functions, always check for asymptotes by analyzing the denominator and limits of the function as xx approaches large values.

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Math Problem Analysis

Mathematical Concepts

Algebra
Functions
Rational Functions
Intercepts
Domain

Formulas

f(x) = (6x^2) / (x^2 + 9)
x-intercepts: set f(x) = 0
y-intercepts: evaluate f(0)
Solving f(x) = 1

Theorems

Rational function domain and intercept properties
Algebraic manipulation to solve for x

Suitable Grade Level

Grades 9-12