Solve g(x)=-1/3(x-5)^2-1

$$g ( x ) = - \frac { 1 } { 3 } ( x - 5 ) ^ { 2 } - 1$$

Evaluate

$$-\frac{\left(x-5\right)^{2}}{3}-1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-\frac{1}{3}\left(x^{2}-10x+25\right)-1$$

Use the distributive property to multiply -\frac{1}{3} by x^{2}-10x+25.

$$-\frac{1}{3}x^{2}+\frac{10}{3}x-\frac{25}{3}-1$$

Subtract 1 from -\frac{25}{3} to get -\frac{28}{3}.

$$-\frac{1}{3}x^{2}+\frac{10}{3}x-\frac{28}{3}$$

Expand

$$\frac{-x^{2}+10x-28}{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$-\frac{1}{3}\left(x^{2}-10x+25\right)-1$$

Use the distributive property to multiply -\frac{1}{3} by x^{2}-10x+25.

$$-\frac{1}{3}x^{2}+\frac{10}{3}x-\frac{25}{3}-1$$

Subtract 1 from -\frac{25}{3} to get -\frac{28}{3}.

$$-\frac{1}{3}x^{2}+\frac{10}{3}x-\frac{28}{3}$$