Question

The axis of symmetry for the graph of the function f start bracket x end bracket equals one-quarter x square plus b x plus 10 is x=6. What is the value of b?

Answer 1

Value of b = -3 if The axis of symmetry for the graph of the function f(x) = x²/4  + bx  + 10  is 6

Step-by-step explanation:

f(x) = x²/4  + bx  + 10

The axis of symmetry for the graph of the function is 6

so root would be

6 + a  & 6 - a

Sum of Roots = 6  + a + 6 - a = 12

Sum of roots  = - b/a  =   -b/(1/4)  = - 4b

-4b = 12

=> b = -3

Value of b = -3

Value of b = -3 if The axis of symmetry for the graph of the function f(x) = x²/4  + bx  + 10  is 6

Learn more:

If one root of the quadratic equation kx^2-3x-1=0 is 1/2 the find value ...

https://brainly.in/question/7551719

If one root of the quadratic equation 2x^2+kx-6=0 is 2 then fin find ...

https://brainly.in/question/7564820

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

f(x) = x²/4 + bx + 10

The axis of symmetry for the graph of the function is 6

so root would be

6 + a & 6 - a

Sum of Roots = 6 + a + 6 - a = 12

Sum of roots = - b/a = -b/(1/4) = - 4b

-4b = 12

=> b = -3

Value of b = -3

Value of b = -3 if The axis of symmetry for the graph of the function f(x) = x²/4 + bx + 10 is 6