Question

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

Answer 1

Answer:

Step-by-step explanation:

f(x) = 1/4 x²+bx+10

the derivate is : f'(x) = 1/2 x +b

you have : f'(6)=0

1/2 (6)+b=0

3+b =0

b = -3

so f(x) = 1/4 x²-3x+10.......f(6) =1/4(6)² -3(6)+10 =9-18+10 =1

f(x) = 1/4(x-6)²+1... the vertex form

Answer 2

Value of 'b' from the quadratic function will be (-3).

Axis of symmetry of a parabola,

If the quadratic equation for the parabolic path has been given as,

• f(x) = ax² + bx + c

Axis of symmetry of the parabola will be given by,

Axis of symmetry = $-\frac{b}{2a}$

  Quadratic function given in the question → $f(x)=\frac{1}{4}x^2+bx+10$

Compare this equation by $f(x)=ax^2+bx+c$

$a=\frac{1}{4},b=b,c=10$

Axis of symmetry of the parabola will be,

Axis of symmetry = $-\frac{b}{2(0.25)}$

                             = $-2b$

Axis of symmetry has been given as x = 6,

$6=-2b$

$b=-3$

    Therefore, value of b will be (-3).

Learn more about the quadratic function here,

https://brainly.com/question/1435393?referrer=searchResults