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What is the vertex of f(x) = x^{2} - 6x + 10?

User Zkolnik
by
8.4k points

1 Answer

1 vote

Answer:

(3, 1)

Explanation:

First, find the equation of the axis of symmetry, which passes vertically through the vertex. The coefficients of f(x) = x^{2} - 6x + 10 are a = 1, b = -6 and c = 10. The formula for the axis of symmetry is

-b -(-6)

x = ------- and in this particular case, x = -------- = 3

2a 2(1)

Next, use this x-vaue (3) as the divisor in synthetic division. The value of the remainder represents the y coordinate of the vertex.

3 1 -6 10

3 -9

-------------------

1 -3 1

Thus, the coordinates of the vertex are (3, 1).

User Colonel Beauvel
by
7.6k points

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