Question

the axis of symmetry for the graph of the function f(x) =3/4 3x2 bx 4 is x =3/2 . what is the value of b?

Answer 1

Hello,
y=3x²+bx+3/4+4
= 3(x²+2b/6*x+b²/36)+19/4-b²/12
=3(x+b/6)²+19/4-b²/12
==>-b/6=3/2==>b=-9

Answer 2

Answer:

The value of b is -9.

Step-by-step explanation:

We are given a equation as:

$f(x)=\frac{3}{4}+3x^2+bx+4$

$f(x)=3x^2+bx+\frac{19}{4}$

clearly the graph of this function will be a parabola.

for any quadratic equation of the type $f(x)=ax^2+bx+c$

the equation of axis of symmetry is given by:

$x=\frac{-b}{2a}$  , here a=3 and b=b

also we are given axis of symmetry as $x=\frac{3}{2}$

that means $\frac{-b}{2\times3}=\frac{3}{2}$

$b=-9$