Question

Consider the function f(x)=x^2+bx−21, where b is a constant. If the function has an axis of symmetry at x=2, what is the value of b?

Answer 1

2x+b

Explanation:

f(x)=x

2

+bx−21

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax

n

is nax

n−1

.

2x

2−1

+bx

1−1

Subtract 1 from 2.

2x

1

+bx

1−1

Subtract 1 from 1.

2x

1

+bx

0

For any term t, t

1

=t.

2x+bx

0

For any term t except 0, t

0

=1.

2x+b×1

For any term t, t×1=t and 1t=t.

2x+b