Math Problem Statement

f(x) = 3x^2 - x - 2. (a) Is the point (2,8) on the graph of f? (b) If x = -1, what is f(x)? What point is on the graph of f? (c) If f(x) = -2, what is x? What point(s) are on the graph of f? (d) What is the domain of f? (e) List the x-intercept(s), if any, of the graph of f. (f) List the y-intercept, if any, of the graph of f.

Solution

Let's go through each question step by step based on the given function f(x)=3x2x2f(x) = 3x^2 - x - 2:

(a) Is the point (2,8) on the graph of ff?

To check if the point (2, 8) lies on the graph, we substitute x=2x = 2 into f(x)f(x) and see if f(2)=8f(2) = 8.

f(2)=3(2)222=3(4)22=1222=8f(2) = 3(2)^2 - 2 - 2 = 3(4) - 2 - 2 = 12 - 2 - 2 = 8

Since f(2)=8f(2) = 8, the point (2, 8) 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)=3(1)2(1)2=3(1)+12=3+12=2f(-1) = 3(-1)^2 - (-1) - 2 = 3(1) + 1 - 2 = 3 + 1 - 2 = 2

Thus, f(1)=2f(-1) = 2, so the point (1,2)(-1, 2) is on the graph of ff.

(c) If f(x)=2f(x) = -2, what is xx? What point(s) are on the graph of ff?

We solve the equation f(x)=2f(x) = -2:

3x2x2=23x^2 - x - 2 = -2

Simplifying:

3x2x=03x^2 - x = 0

Factor the equation:

x(3x1)=0x(3x - 1) = 0

Thus, x=0x = 0 or 3x1=03x - 1 = 0, which gives x=13x = \frac{1}{3}.

Therefore, the points are (0,2)(0, -2) and (13,2)\left( \frac{1}{3}, -2 \right).

(d) What is the domain of ff?

The function f(x)=3x2x2f(x) = 3x^2 - x - 2 is a quadratic function, and quadratic functions are defined for all real numbers. Therefore, the domain of ff is all real numbers or (,)(-\infty, \infty).

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

To find the x-intercepts, set f(x)=0f(x) = 0 and solve for xx:

3x2x2=03x^2 - x - 2 = 0

We can solve this quadratic using the quadratic formula:

x=(1)±(1)24(3)(2)2(3)x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-2)}}{2(3)}

Simplifying:

x=1±1+246=1±256=1±56x = \frac{1 \pm \sqrt{1 + 24}}{6} = \frac{1 \pm \sqrt{25}}{6} = \frac{1 \pm 5}{6}

Thus:

x=1+56=1orx=156=23x = \frac{1 + 5}{6} = 1 \quad \text{or} \quad x = \frac{1 - 5}{6} = -\frac{2}{3}

The x-intercepts are (1,0)(1, 0) and (23,0)\left( -\frac{2}{3}, 0 \right).

(f) List the y-intercept, if any, of the graph of ff.

The y-intercept occurs when x=0x = 0. Substituting x=0x = 0 into the function:

f(0)=3(0)202=2f(0) = 3(0)^2 - 0 - 2 = -2

Thus, the y-intercept is (0,2)(0, -2).


Would you like further details on any of these steps, or any other explanations?

Related Questions:

  1. How can we use the vertex form to find the vertex of this quadratic function?
  2. What is the axis of symmetry for the function f(x)=3x2x2f(x) = 3x^2 - x - 2?
  3. How do you determine if the parabola opens upward or downward?
  4. How would you find the maximum or minimum value of f(x)f(x)?
  5. Can you determine the intervals where the function is increasing or decreasing?

Tip:

When solving quadratic equations, the discriminant b24acb^2 - 4ac can tell you if there are 0, 1, or 2 real solutions (x-intercepts).

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Algebra
Coordinate Geometry

Formulas

Quadratic function formula: f(x) = ax^2 + bx + c
Quadratic equation: ax^2 + bx + c = 0
Quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a

Theorems

Quadratic Formula
Domain of Polynomial Functions
X-Intercept and Y-Intercept Calculation

Suitable Grade Level

Grades 9-12