Math Problem Statement

f(x)=−x2+6x−10f(x)=-x2+6x-10 Write the Vertex as an Ordered Pair

Solution

To find the vertex of the quadratic function f(x)=x2+6x10f(x) = -x^2 + 6x - 10, we can use the vertex formula. For a function in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, the x-coordinate of the vertex is given by:

x=b2ax = -\frac{b}{2a}

In this case:

  • a=1a = -1
  • b=6b = 6
  1. Calculate the x-coordinate of the vertex:

x=621=62=3x = -\frac{6}{2 \cdot -1} = -\frac{6}{-2} = 3

  1. Substitute x=3x = 3 back into the function to find the y-coordinate of the vertex:

f(3)=(3)2+6310f(3) = -(3)^2 + 6 \cdot 3 - 10 =9+1810= -9 + 18 - 10 =1= -1

Thus, the vertex of the function is:

(3,1)(3, -1)

The vertex as an ordered pair is (3,1)(3, -1).

Would you like a breakdown of the steps or have any questions?

Here are five related questions for further practice:

  1. How would you find the vertex of g(x)=2x28x+5g(x) = 2x^2 - 8x + 5?
  2. What is the effect of the coefficient aa on the orientation of the parabola?
  3. How can you determine if the vertex is a maximum or minimum point?
  4. How would the vertex formula apply if f(x)=x24x+4f(x) = x^2 - 4x + 4?
  5. How does changing the constant term cc affect the vertex?

Tip: For any quadratic function, the vertex can be thought of as the "turning point" of the parabola.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Vertex Form

Formulas

Vertex formula: x = -b / 2a
Quadratic function: f(x) = ax^2 + bx + c

Theorems

Vertex theorem for parabolas

Suitable Grade Level

Grades 8-10