Question

if $f(x)=3x^2 + 12x + 9$, what are the coordinates of the vertex?
a. $(0,9)$
b. $(-3,6)$
c. $(-2,-3)$
d. $(1,4)$

Explanation:

Step1: Recall vertex formula for quadratic

For \( f(x)=ax^2 + bx + c \), vertex x - coordinate is \( x = -\frac{b}{2a} \). Here, \( a = 3 \), \( b = 12 \).
\( x = -\frac{12}{2\times3} = -2 \)

Step2: Find y - coordinate

Substitute \( x = -2 \) into \( f(x) \): \( f(-2)=3(-2)^2 + 12(-2)+9 \)
\( = 3\times4 - 24 + 9 = 12 - 24 + 9 = -3 \)
So vertex is \( (-2, -3) \).

Answer:

c. (-2, -3)