Question

the axis of symmetry for the graph of the function is f(x) = 1/4x2 bx 10 is x = 6. what is the value of b?

Answer 1

Hello,

y=x²/4+bx+10=1/4(x²+2*2b+4b²)+10-b²
=(1/4'x+2b)²+10-b²
==>-2b=6==>b=-3

Answer 2

we have

`f(x)=(1)/(4) x^(2) +bx+10`

This is a vertical parabola open upward

The axis of symmetry is equal to the x-coordinate of the vertex

The vertex is the point (h,k)

the equation of the axis of symmetry is
`x=h`

In this problem

`x=6`

so the x-coordinate of the vertex is
`h=6`

Convert the quadratic equation in vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)-10=(1)/(4) x^(2) +bx`

Factor the leading coefficient

`f(x)-10=(1)/(4)(x^(2) +4bx)`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)-10+b^(2)=(1)/(4)(x^(2) +4bx+4b^(2))`

Rewrite as perfect squares

`f(x)+(b^(2)-10)=(1)/(4)(x+2b)^(2)`

`f(x)=(1)/(4)(x+2b)^(2)-(b^(2)-10)`

remember that

`h=6`

so

`(x+2b)=(x-6)`

`2b=-6`

`b=-3`

substitute in the equation

`f(x)=(1)/(4)(x-6)^(2)-((-3)^(2)-10)`

`f(x)=(1)/(4)(x-6)^(2)+1`

therefore

the answer is

`b=-3`