The function f is defined by f(x) = x^2 - 10x + 29. For what value of x does f(x) reach it… — SAT

The function $f$ is defined by $f(x) = x^2 - 10x + 29$ . For what value of $x$ does $f(x)$ reach its minimum value?

Answer: 5

The function reaches its minimum value at

x = 5

A quadratic function defined by

f(x) = ax^2 + bx + c

where

a > 0

reaches its minimum value at its vertex. The

x

-coordinate of the vertex is calculated using the formula

x = -\frac{b}{2a}

. For

f(x) = x^2 - 10x + 29

, the coefficients are

a = 1

and

b = -10

. Substituting these values into the formula gives

x = -\frac{-10}{2(1)} = 5

Identify the coefficients of the quadratic function in standard form

f(x) = ax^2 + bx + c

For

f(x) = x^2 - 10x + 29

, the coefficients are

a = 1

and

b = -10

To use the vertex formula

x = -\frac{b}{2a}

Apply the vertex formula to calculate the

x

-coordinate where the function reaches its minimum.

x = 5

Since the coefficient of

x^2

is positive (

a = 1

), the parabola opens upward, and its minimum value occurs at the vertex.

Finding the x-coordinate of the vertex of a quadratic function

Question