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

we have

$f(x)=\frac{1}{4} x^{2} +bx+10$

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

$f(x)-10=\frac{1}{4} x^{2} +bx$

Factor the leading coefficient

$f(x)-10=\frac{1}{4}(x^{2} +4bx)$

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

$f(x)-10+b^{2}=\frac{1}{4}(x^{2} +4bx+4b^{2})$

Rewrite as perfect squares

$f(x)+(b^{2}-10)=\frac{1}{4}(x+2b)^{2}$

$f(x)=\frac{1}{4}(x+2b)^{2}-(b^{2}-10)$

remember that

$h=6$

so

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

$2b=-6$

$b=-3$

substitute in the equation

$f(x)=\frac{1}{4}(x-6)^{2}-((-3)^{2}-10)$

$f(x)=\frac{1}{4}(x-6)^{2}+1$

therefore

the answer is

$b=-3$

Answer 2

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