## Calculus 8th Edition

(a) $8a-6a^2$ (b)For point (1,5) $y=2x+3$ For point (2,3) $y=-8x+19$ (c) red = $y=3+4x^2-2x^3$ blue = $y=2x+3$ green = $y=-8x+19$ (a) $y=3+4x^2-2x^3$ at the point where $x=a$ Step 1: The question is asking for the slope of the tangent line so we have to find the first derivative. We can find the first derivative by using the power rule. $y'=8x-6x^2$ Step 2: Plug in $a$ wherever you see an $x$ in the equation. $8a-6a^2$ (b) points (1,5) and (2.3) Step 1: Plug in the given x values into $y'=8x-6x^2$. $y'(1)=8(1)-6(1)^2$ $y'(1)=2$ $y'(2)=8(2)-6(2)^2$ $y'(2)=-8$ Step 2: Now that we have calculated the slope of the tangent lines, we can write the equation of the tangent line in point slope form. For point (1,5) $y-5=2(x-1)$ For point (2,3) $y-3=-8(x-2)$ Step 3: Rewrite the equations into standard form. For point (1,5) $y=2x+3$ For point (2,3) $y=-8x+19$ (c) Plug the original function and equations of the tangent lines into a graphing calculator or graph by hand. red = $y=3+4x^2-2x^3$ blue = $y=2x+3$ green = $y=-8x+19$