Consider the function y=g(x)=-x^2+5x+2 (a) Use the limit definition to compute a formula for y=g'(x) y= (b) Determine the slope of the tangent line to y=g(x) at the value x=2 slope= (c) Compute g(2) g(2)= (d) The equation for the tangent line to y=g(x) at the point (2,g(2)) written in point-slope form, is y-square =square (x-square ) Fill in the blanks with the appropriate numbers.

(a) To compute the derivative of the function $y=g(x)=-x^{2}+5x+2$ using the limit definition, we need to find the limit of the difference quotient as the change in $x$ approaches 0.<br /><br />The difference quotient is given by:<br />$\frac{g(x+h)-g(x)}{h}$<br /><br />Substituting the function $g(x)$ into the difference quotient, we get:<br />$\frac{(-x^2+5x+2)-(x^2-5x-2)}{h} = \frac{-2x^2+10x+4}{h}$<br /><br />Taking the limit as $h$ approaches 0, we get:<br />$\lim_{h\to 0} \frac{-2x^2+10x+4}{h} = -2x+5$<br /><br />Therefore, the formula for the derivative $y=g'(x)$ is:<br />$y=-2x+5$<br /><br />(b) To determine the slope of the tangent line to $y=g(x)$ at the value $x=2$, we need to evaluate the derivative $g'(x)$ at $x=2$.<br /><br />Substituting $x=2$ into the derivative formula, we get:<br />$g'(2)=-2(2)+5=1$<br /><br />Therefore, the slope of the tangent line to $y=g(x)$ at the value $x=2$ is 1.<br /><br />(c) To compute $g(2)$, we need to substitute $x=2$ into the original function $y=g(x)=-x^{2}+5x+2$.<br /><br />Substituting $x=2$, we get:<br />$g(2)=-2^2+5(2)+2=8$<br /><br />Therefore, $g(2)=8$.<br /><br />(d) The equation for the tangent line to $y=g(x)$ at the point $(2,g(2))$ written in point-slope form is:<br />$y-g(2)=g'(2)(x-2)$<br /><br />Substituting the values of $g(2)$ and $g'(2)$, we get:<br />$y-8=1(x-2)$<br /><br />Therefore, the equation for the tangent line to $y=g(x)$ at the point $(2,g(2))$ written in point-slope form is:<br />$y-8=1(x-2)$