Math Problem Statement

For 𝑓(π‘₯) = βˆ’π‘₯2 + 6π‘₯ + 7 , find: a) vertex ________________ b) axis of symmetry ________ c) x-intercepts ____________ d) y-intercept ___________ e) domain ________ f) range

Solution

For the quadratic function f(x)=βˆ’x2+6x+7f(x) = -x^2 + 6x + 7, let's break down each part:

a) Vertex

The vertex of a parabola in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c can be found using the formula: x=βˆ’b2ax = \frac{-b}{2a} Here, a=βˆ’1a = -1, b=6b = 6, and c=7c = 7. Plugging in the values: x=βˆ’62(βˆ’1)=3x = \frac{-6}{2(-1)} = 3 Now, substitute x=3x = 3 into the original function to find the y-coordinate of the vertex: f(3)=βˆ’(3)2+6(3)+7=βˆ’9+18+7=16f(3) = -(3)^2 + 6(3) + 7 = -9 + 18 + 7 = 16 So, the vertex is: (3,16)\boxed{(3, 16)}

b) Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. Since the x-coordinate of the vertex is x=3x = 3, the axis of symmetry is: x=3\boxed{x = 3}

c) X-intercepts

The x-intercepts are the points where f(x)=0f(x) = 0, so we solve: βˆ’x2+6x+7=0-x^2 + 6x + 7 = 0 Multiply the equation by βˆ’1-1 to simplify: x2βˆ’6xβˆ’7=0x^2 - 6x - 7 = 0 Now, solve this quadratic using the quadratic formula: x=βˆ’(βˆ’6)Β±(βˆ’6)2βˆ’4(1)(βˆ’7)2(1)=6Β±36+282=6Β±642x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-7)}}{2(1)} = \frac{6 \pm \sqrt{36 + 28}}{2} = \frac{6 \pm \sqrt{64}}{2} x=6Β±82x = \frac{6 \pm 8}{2} Thus, the solutions are: x=6+82=7andx=6βˆ’82=βˆ’1x = \frac{6 + 8}{2} = 7 \quad \text{and} \quad x = \frac{6 - 8}{2} = -1 So, the x-intercepts are: (βˆ’1,0) and (7,0)\boxed{(-1, 0) \text{ and } (7, 0)}

d) Y-intercept

The y-intercept occurs when x=0x = 0. Substituting x=0x = 0 into the function: f(0)=βˆ’(0)2+6(0)+7=7f(0) = -(0)^2 + 6(0) + 7 = 7 So, the y-intercept is: (0,7)\boxed{(0, 7)}

e) Domain

The domain of a quadratic function is always all real numbers because the function is defined for every xx. Thus, the domain is: (βˆ’βˆž,∞)\boxed{(-\infty, \infty)}

f) Range

Since the parabola opens downwards (because a=βˆ’1a = -1 is negative), the maximum value of the function occurs at the vertex. The maximum value is the y-coordinate of the vertex, which is 16. Therefore, the range is: (βˆ’βˆž,16]\boxed{(-\infty, 16]}


Would you like a more detailed explanation of any part?

Here are 5 related questions to deepen your understanding:

  1. How do you find the vertex form of a quadratic function?
  2. How can you determine whether a parabola opens upwards or downwards?
  3. How do you find the discriminant of a quadratic equation and what does it tell you?
  4. How is the quadratic formula derived?
  5. How can the concept of symmetry help in solving quadratic equations graphically?

Tip: Always check the sign of the quadratic coefficient (a) to determine the direction of the parabola (upward or downward).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Parabolas

Formulas

Vertex formula: x = -b/(2a)
Quadratic formula: x = [-b ± √(b^2 - 4ac)] / 2a
Intercept calculations

Theorems

The vertex theorem for parabolas
The quadratic formula theorem

Suitable Grade Level

Grades 8-10