Question

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

Answer 1

So.. hmmm notice the vertex location for a parabola
is located at $\bf y = ax^2+bx+c \qquad \\ \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad f\left(-\cfrac{{{ b}}}{2{{ a}}}\right)\right)$

also, notice the axis of symmetry in the picture, x = 6
the axis of symmetry is located at the same x-value as
the vertex's x-coordinate, so whatever$\bf -\cfrac{{{ b}}}{2{{ a}}}$ is
is equals to 6, since we know, by the axis of symmetry, 
that that x spot IS indeed 6

so one can say $\bf y = 1x^2+bx+10 \\\\ -\cfrac{{{ b}}}{2{{ a}}}\implies -\cfrac{{{ b}}}{2{{ (1)}}} \\ \quad \\\\ and\qquad -\cfrac{{{ b}}}{2{{ (1)}}}=6\impliedby \textit{ solve for "b"}$

Answer 2

Answer:

$b=-12$

Step-by-step explanation:

we have

$f(x)=x^{2}+bx+10$

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex of the parabola

and the axis of symmetry is equal to $x=h$

In this problem we have the axis of symmetry $x=6$

so

the x-coordinate of the vertex is equal to $6$  

therefore

For $x=6+1=7$ -----> one unit to the right of the vertex

Find the value of $f(7)$

$f(7)=7^{2}+b(7)+10$

$f(7)=59+7b$

For $x=6-1=5$ -----> one unit to the left of the vertex

Find the value of $f(5)$

$f(5)=5^{2}+b(5)+10$

$f(5)=35+5b$

Remember that

$f(5)=f(7)$ ------> the x-coordinates are at the same distance from the axis of symmetry

so

$59+7b=35+5b$ ------> solve for b

$7b-5b=35-59$

$2b=-24$

$b=-12$