Question

score: 0 of 1 point
determine the vertex of the quadratic function $y = x^2 - 10x + 21$.
a $(5, 42)$
b $(10, -4)$
c $(5, 21)$
d $(5, -4)$
e $(10, 21)$

Explanation:

Step1: Find the x - coordinate of the vertex

For a quadratic function \(y = ax^{2}+bx + c\), the formula for the x - coordinate of the vertex is \(x=-\frac{b}{2a}\).
In the function \(y=x^{2}-10x + 21\), \(a = 1\) and \(b=-10\).
So, \(x=-\frac{-10}{2\times1}=5\).

Step2: Find the y - coordinate of the vertex

Substitute \(x = 5\) into the function \(y=x^{2}-10x + 21\).
\(y=(5)^{2}-10\times5 + 21=25-50 + 21=-4\).

Answer:

D. \((5,-4)\)