Solve f(x)=-1/3(x-6)^2-3

$$f ( x ) = - 1 / 3 ( x - 6 ) ^ { 2 } - 3$$

Evaluate

$$-\frac{\left(x-6\right)^{2}}{3}-3$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$-\frac{1}{3}\left(x^{2}-12x+36\right)-3$$

Use the distributive property to multiply -\frac{1}{3} by x^{2}-12x+36.

$$-\frac{1}{3}x^{2}+4x-12-3$$

Subtract 3 from -12 to get -15.

$$-\frac{1}{3}x^{2}+4x-15$$

Expand

$$-\frac{x^{2}}{3}+4x-15$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$-\frac{1}{3}\left(x^{2}-12x+36\right)-3$$

Use the distributive property to multiply -\frac{1}{3} by x^{2}-12x+36.

$$-\frac{1}{3}x^{2}+4x-12-3$$

Subtract 3 from -12 to get -15.

$$-\frac{1}{3}x^{2}+4x-15$$