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

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

Evaluate

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

Solution Steps

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

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

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

$$-\frac{1}{2}x^{2}+5x-\frac{25}{2}-1$$

Subtract 1 from -\frac{25}{2} to get -\frac{27}{2}.

$$-\frac{1}{2}x^{2}+5x-\frac{27}{2}$$

Expand

$$-\frac{x^{2}}{2}+5x-\frac{27}{2}$$

Solution Steps

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

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

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

$$-\frac{1}{2}x^{2}+5x-\frac{25}{2}-1$$

Subtract 1 from -\frac{25}{2} to get -\frac{27}{2}.

$$-\frac{1}{2}x^{2}+5x-\frac{27}{2}$$