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Question
question 6 of 10
what is the vertex of the graph of the function below?
y = x² - 10x + 21
a. (5, 0)
b. (3, -4)
c. (5, -4)
d. (3, 0)
Step1: Recall vertex formula for quadratic
For \( y = ax^2 + bx + c \), x - coordinate of vertex is \( x = -\frac{b}{2a} \). Here, \( a = 1 \), \( b = -10 \).
\( x = -\frac{-10}{2\times1} = 5 \)
Step2: Find y - coordinate
Substitute \( x = 5 \) into \( y = x^2 - 10x + 21 \).
\( y = 5^2 - 10\times5 + 21 = 25 - 50 + 21 = -4 \)
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C. (5, -4)