Question

Which function in vertex form is equivalent to f(x) = x2 + x +1? f(x) = (x + one-quarter) squared + three-quarters f(x) = (x + one-quarter) squared + five-quarters f(x) = (x + one-half) squared + three-quarters f(x) = (x + one-half) squared + five-quarters

Answer 1

The function in vertex form is $f(x)=(x+\frac{1}{2})^{2}+\frac{3}{4}$ ⇒ 3rd answer

Step-by-step explanation:

The vertex form of the quadratic function f(x) = ax² + bx + c is

f(x) = a(x - h)² + k, where

• a is the coefficient of x²
• (h , k) are the coordinates of the vertex point
• $h=\frac{-b}{2a}$ , wher b is the coefficient of x
• k = f(h), that means value f(x) when x = h

∵ f(x) = x² + x + 1

∴ a = 1 , b = 1

∵ $h=\frac{-b}{2a}$

- Substitute the values of a and b to find h

∴ $h=\frac{-1}{2(1)}$

∴ $h=\frac{-1}{2}$

Substitute the value of x in f(x) by the value of h to find k

∵ f( $\frac{-1}{2}$ ) = $(\frac{-1}{2})^{2}+\frac{-1}{2}+1$

∴ f( $\frac{-1}{2}$ ) = $\frac{1}{4}-\frac{1}{2}+1$

∴ f( $\frac{-1}{2}$ ) = $\frac{3}{4}$

- k is the value of f(x) when x = h

∵ h = $\frac{-1}{2}$

∴ k = f( $\frac{-1}{2}$ )

∴ k = $\frac{3}{4}$

Substitute the values of a, h and k in the vertex form

∵ f(x) = a(x - h)² + k

∵ a = 1 , $h=\frac{-1}{2}$ , $k=\frac{3}{4}$

∴ $f(x)=1(x-\frac{-1}{2})^{2}+\frac{3}{4}$

∴ $f(x)=(x+\frac{1}{2})^{2}+\frac{3}{4}$

The function in vertex form is $f(x)=(x+\frac{1}{2})^{2}+\frac{3}{4}$

Learn more:

You can learn more about the quadratic functions in brainly.com/question/9390381

#LearnwithBrainly

Answer 2

Answer:

Its C on edg

Step-by-step explanation:

follow my ifunny "dankmemehistory"