Question

for the quadratic function $f(x)=x^2 - 10x + 25$, what is the $y$-coordinate of the vertex?
a. -5
b. 0
c. 1
d. 5

Explanation:

Step1: Recall vertex formula for quadratic

For \( f(x)=ax^2 + bx + c \), vertex x - coordinate is \( x = -\frac{b}{2a} \). Here, \( a = 1 \), \( b=-10 \), so \( x = -\frac{-10}{2\times1}=5 \).

Step2: Find y - coordinate

Substitute \( x = 5 \) into \( f(x)=x^2 - 10x + 25 \). \( f(5)=5^2-10\times5 + 25=25 - 50 + 25 = 0 \).

Answer:

b. 0