Question

the axis of symmetry for the graph of the function ( f(x)=\frac{1}{4}x^2 + bx + 10 ) is ( x = 6 ). what is the value of ( b )?
( circ -12 )
( circ -3 )
( circ \frac{1}{2} )
( circ 3 )

Explanation:

Step1: Recall the axis of symmetry formula for a quadratic function.

For a quadratic function in the form \( f(x) = ax^2 + bx + c \), the axis of symmetry is given by the formula \( x = -\frac{b}{2a} \).

Step2: Identify the values of \( a \) and the axis of symmetry.

In the given function \( f(x) = \frac{1}{4}x^2 + bx + 10 \), we have \( a = \frac{1}{4} \) and the axis of symmetry \( x = 6 \).

Step3: Substitute the known values into the axis of symmetry formula and solve for \( b \).

Substitute \( a = \frac{1}{4} \) and \( x = 6 \) into \( x = -\frac{b}{2a} \):

$$ 6 = -\frac{b}{2 \times \frac{1}{4}} $$

Simplify the denominator: \( 2 \times \frac{1}{4} = \frac{1}{2} \)
So the equation becomes:

$$ 6 = -\frac{b}{\frac{1}{2}} $$

Which is equivalent to:

$$ 6 = -2b $$

Solve for \( b \) by dividing both sides by -2:

$$ b = \frac{6}{-2} = -3 $$

Answer:

-3