Question

Answer the questions about each function. $$ f(x)=-3 x^{2}+5 x $$ (a) Is the point (-1,2) on the graph of \(f ?\) (b) If \(x=-2,\) 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 \(x\) -intercepts, if any, of the graph of \(f\). (f) List the \(y\) -intercept, if there is one, of the graph of \(f\).

Short answer

a) No. b) -22, (-2, -22). c) \( x = \frac{1}{3} \) or 2, \( \left( \frac{1}{3}, -2 \right) \) and (2, -2). d) All real numbers. e) (0, 0) and \( \left( \frac{5}{3}, 0 \right) \). f) (0,0).

Step-by-step solution

- Verify if Point (-1,2) is on the Graph

To determine if the point (-1,2) is on the graph, substitute x = -1 into the function and see if the result equals 2. Calculate \[ f(-1) = -3(-1)^2 + 5(-1) \] \[ = -3(1) - 5 \] \[ = -3 - 5 \] \[ = -8 \] Since \( f(-1) \) is not equal to 2, the point (-1,2) is not on the graph.

- Calculate f(x) for x = -2

Substitute x = -2 into the function to find \( f(-2) \). Calculate \[ f(-2) = -3(-2)^2 + 5(-2) \] \[ = -3(4) - 5(2) \] \[ = -12 - 10 \] \[ = -22 \] So, the point (-2, -22) is on the graph.

- Solve f(x) = -2

Set the function equal to -2 and solve for x. \[ -3x^2 + 5x = -2 \] \[ -3x^2 + 5x + 2 = 0 \] Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -3, b = 5, c = 2 \). \[ x = \frac{-5 \pm \sqrt{5^2 - 4(-3)(2)}}{2(-3)} \] \[ = \frac{-5 \pm \sqrt{25 + 24}}{-6} \] \[ = \frac{-5 \pm \sqrt{49}}{-6} \] \[ = \frac{-5 \pm 7}{-6} \] \[ x = \frac{2}{-6} \] or \( x = \frac{-12}{-6} \) \[ = \frac{1}{3} \] or \( 2 \) So the points \( \left( \frac{1}{3}, -2 \right) \) and (2, -2) are on the graph.

- Determine the Domain of f(x)

The function \( f(x) = -3x^2 + 5x \) is a polynomial, and polynomials are defined for all real numbers. So, the domain is all real numbers \( (-\infty, \infty) \).

- Find x-Intercepts

Set \( f(x) = 0 \) to find x-intercepts. \[ -3x^2 + 5x = 0 \] Factor out x: \[ x(-3x + 5) = 0 \] Thus, \( x = 0 \) or \( -3x + 5 = 0 \) \[ x = \frac{5}{3} \] So, the x-intercepts are (0, 0) and \( \left( \frac{5}{3}, 0 \right) \).

- Find y-Intercept

Substitute \( x = 0 \) to find the y-intercept: \[ f(0) = -3(0)^2 + 5(0) \] \[ = 0 \] So, the y-intercept is (0, 0).

Key concepts

Graphing Quadratic Functions

Quadratic functions are defined by the equation \( f(x) = ax^2 + bx + c \). The graph of a quadratic function is a parabola. If the coefficient \( a \) is positive, the parabola opens upwards; if \( a \) is negative, it opens downwards. The vertex of the parabola, which is its maximum or minimum point, can be found using the formula \( x = -\frac{b}{2a} \). For our function \( f(x) = -3x^2 + 5x \), the parabola opens downwards because \( a = -3 \) is negative. Graphing a quadratic function involves finding key points, such as the vertex, x-intercepts, and y-intercept. Plot these points and draw a smooth curve through them to complete the graph.

Finding X-Intercepts

To find the x-intercepts of a quadratic function, set \( f(x) \) to 0 and solve the quadratic equation. For the function \( f(x) = -3x^2 + 5x \), set \( -3x^2 + 5x = 0 \). Factoring out \( x \) gives \( x(-3x + 5) = 0 \). This results in two solutions: \( x = 0 \) and \( x = \frac{5}{3} \). Therefore, the x-intercepts are at points \( (0,0) \) and \( \left( \frac{5}{3}, 0 \right) \). X-intercepts are essential as they show where the graph crosses the x-axis. These points are also known as the roots or solutions of the equation.

Finding Y-Intercepts

Finding the y-intercept of a quadratic function involves setting \( x = 0 \) and calculating \( f(0) \). This gives the point at which the graph crosses the y-axis. For our function \( f(x) = -3x^2 + 5x \), substituting \( x = 0 \) results in \( f(0) = 0 \). Therefore, the y-intercept is at point \( (0, 0) \). Intercepts help us understand the behavior of the graph. Note that in this specific case, the y-intercept and one of the x-intercepts are the same point.

Domain of a Function

The domain of a quadratic function refers to all the possible values of \( x \) for which the function is defined. Quadratic functions like \( f(x) = -3x^2 + 5x \) are polynomials, and polynomials are defined for all real numbers. Therefore, the domain of this quadratic function is all real numbers, written as \( (-\infty, \infty) \). This means the graph extends indefinitely in both the positive and negative directions along the x-axis. Understanding the domain is crucial, as it tells us the scope within which the function operates.