Question

question 27 (mandatory) (1 point)
determine which coordinate is the vertex of $f(x) = 4x^2 - 8x + 11$ without graphing the parabola.
\bigcirc a) $(1, 10)$
\bigcirc b) $(1, 12)$
\bigcirc c) $(1, 7)$
\bigcirc d) $(-1, 7)$

Explanation:

Step1: Find x-coordinate of vertex

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a = 4 \), \( b = -8 \). So \( x = -\frac{-8}{2\times4} = \frac{8}{8} = 1 \).

Step2: Find y-coordinate by substitution

Substitute \( x = 1 \) into \( f(x) = 4x^2 - 8x + 11 \). \( f(1) = 4(1)^2 - 8(1) + 11 = 4 - 8 + 11 = 7 \). So the vertex is \( (1, 7) \).

Answer:

c) (1, 7)