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The vertex form of the quadratic function f(x) = -6x² - 60x - 151 is
f(x)= a(x - h)² + k.
What is the value of a?
What is the value of h?
What is the value of k?

Sagot :

theleangreenbean

Vertex Form

This is one way of writing a quadratic function:

[tex]f(x)= a(x-h)^2+k[/tex]

  • (h,k) = vertex
  • a = vertical stretch

Solving the Question

We're given:

[tex]f(x) = -6x^2 - 60x - 151[/tex]

To write a quadratic function in vertex form, we must complete the square.

⇒ Put parentheses around the first two terms containing x and x²:

[tex]f(x) = (-6x^2 - 60x) - 151[/tex]

⇒ Factor out -6 (keeping the x's in the parentheses):

[tex]f(x) = -6(x^2+10x) - 151[/tex]

⇒ To complete the square, add, inside the parentheses, the square of half the coefficient of x.

  • ⇒ In this case, the coefficient of x is 10.
  • ⇒ Half of this value is 5.
  • ⇒ The square of 5 is 25.
  • ⇒ Add this inside the parentheses:

[tex]f(x) = -6(x^2+10x+25) - 151[/tex]

⇒ Now, we cannot randomly introduce a new value into a function. To balance the +25, subtract -6(25) outside the parentheses:

[tex]f(x) = -6(x^2+10x+25) - 151-(-6*25)\\f(x) = -6(x^2+10x+25) - 151-(-150)\\f(x) = -6(x^2+10x+25) - 151+150\\f(x) = -6(x^2+10x+25) - 1[/tex]

⇒ Finally, complete the square.

  • Remember: [tex](a+b)^2=a^2+2ab+b^2[/tex]
  • In this case, a is x, and b is 5:

[tex]f(x) = -6(x+5)^2 - 1[/tex]

Answer

[tex]f(x) = -6(x+5)^2 - 1[/tex]

  • a = -6
  • h = -5
  • h = -1
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