Math Problem Statement

The function f is given by three equivalent quadratic equations: f(x) = x^2 - 2x - 15, f(x) = (x + 3)(x - 5), and f(x) = (x - 1)^2 - 16. Evaluate the following statements about the graph of the function: a) The graph has a y-intercept at (0, 15). b) The coordinates of the vertex are (1, -16). c) The graph of the function opens upward. d) The x-intercepts are at (3, 0) and (-5, 0).

Solution

This image shows a math problem related to quadratic functions, where three equivalent quadratic equations are given, and the task is to evaluate different statements about the graph of the function.

The three quadratic equations are:

  • f(x)=x22x15f(x) = x^2 - 2x - 15
  • f(x)=(x+3)(x5)f(x) = (x + 3)(x - 5)
  • f(x)=(x1)216f(x) = (x - 1)^2 - 16

And the four statements to evaluate are:

a) The graph has a y-intercept at (0,15)(0, 15).

b) The coordinates of the vertex are (1,16)(1, -16).

c) The graph of the function opens upward.

d) The x-intercepts are at (3,0)(3, 0) and (5,0)(-5, 0).

Let me break down each part to help evaluate whether these statements are true or false.


a) The graph has a y-intercept at (0,15)(0, 15).

To check this, we substitute x=0x = 0 in each of the equations to find the y-intercept.

For equation A, f(x)=x22x15f(x) = x^2 - 2x - 15: f(0)=022(0)15=15f(0) = 0^2 - 2(0) - 15 = -15 Thus, the y-intercept is (0,15)(0, -15), not (0,15)(0, 15). So, statement a is false.


b) The coordinates of the vertex are (1,16)(1, -16).

To find the vertex, we need to express the equation in vertex form. Let's check each equation.

For equation A, f(x)=x22x15f(x) = x^2 - 2x - 15, we complete the square to find the vertex form: f(x)=(x1)216f(x) = (x - 1)^2 - 16 This is the vertex form, and the vertex is indeed (1,16)(1, -16). So, statement b is true.


c) The graph of the function opens upward.

A quadratic function opens upward if the coefficient of x2x^2 is positive. In all the given equations, the coefficient of x2x^2 is positive. Therefore, the graph opens upward. So, statement c is true.


d) The x-intercepts are at (3,0)(3, 0) and (5,0)(-5, 0).

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

For equation A, f(x)=x22x15f(x) = x^2 - 2x - 15, we solve: x22x15=0x^2 - 2x - 15 = 0 Factoring gives: (x5)(x+3)=0(x - 5)(x + 3) = 0 Thus, the x-intercepts are x=5x = 5 and x=3x = -3, not 3 and -5. Therefore, statement d is false.


Final Answers:

  • a) False
  • b) True
  • c) True
  • d) False

Would you like me to explain any of these steps in more detail, or do you have any further questions about this problem?


Related Questions:

  1. How do you complete the square to convert a quadratic equation into vertex form?
  2. What does the vertex form of a quadratic function tell you about the graph?
  3. How do you determine whether a parabola opens upward or downward from the equation?
  4. How can you factor a quadratic equation to find the x-intercepts?
  5. How does the discriminant of a quadratic equation help you determine the number of real x-intercepts?

Tip:

When finding the vertex of a quadratic equation, always check if the equation is in standard form ax2+bx+cax^2 + bx + c and then try to complete the square or use the vertex formula x=b2ax = \frac{-b}{2a} to directly find the x-coordinate of the vertex.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Vertex Form
Factoring Quadratics

Formulas

Standard form of a quadratic equation: f(x) = ax^2 + bx + c
Vertex form of a quadratic equation: f(x) = a(x-h)^2 + k
Factoring: f(x) = (x-r1)(x-r2)

Theorems

Properties of parabolas: Opening direction determined by coefficient of x^2
Vertex of a parabola at x = -b/(2a)

Suitable Grade Level

Grade 9-10