Question 128195
*[Tex \LARGE f(x)=4 -x^2] Start with the given function



*[Tex \LARGE f'(x)=-2x] Derive *[Tex \LARGE f(x)=4 -x^2] to get *[Tex \LARGE f'(x)=-2x]


Now to find the slope of the tangent line at (1,3), simply plug in {{{x=1}}} into *[Tex \LARGE f'(x)=-2x]



*[Tex \LARGE f'(1)=-2(1)] Plug in {{{x=1}}}



*[Tex \LARGE f'(1)=-2] Multiply



So the slope of the tangent line at (1,3) is -2 (note: this is where you made a mistake)






Now let's use the Point-Slope Formula to find the equation of the tangent line 



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-3=(-2)(x-1)}}} Plug in {{{m=-2}}}, {{{x[1]=1}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=-2x+(-2)(-1)}}} Distribute {{{-2}}}


{{{y-3=-2x+2}}} Multiply {{{-2}}} and {{{-1}}} to get {{{2}}}


{{{y=-2x+2+3}}} Add 3 to  both sides to isolate y


{{{y=-2x+5}}} Combine like terms {{{2}}} and {{{3}}} to get {{{5}}} 




So the equation of the tangent line is {{{y=-2x+5}}}




Now simply negate and find the reciprocal of the slope {{{m=-2}}} to get {{{m=1/2}}} to get the perpendicular slope



Now let's use the Point-Slope Formula to find the equation of the normal line note: the normal line has a slope of {{{m=1/2}}} and goes through (1,3)



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-3=(1/2)(x-1)}}} Plug in {{{m=1/2}}}, {{{x[1]=1}}}, and {{{y[1]=3}}} (these values are given)



{{{y-3=(1/2)x+(1/2)(-1)}}} Distribute {{{1/2}}}


{{{y-3=(1/2)x-1/2}}} Multiply {{{1/2}}} and {{{-1}}} to get {{{-1/2}}}


{{{y=(1/2)x-1/2+3}}} Add 3 to  both sides to isolate y


{{{y=(1/2)x+5/2}}} Combine like terms {{{-1/2}}} and {{{3}}} to get {{{5/2}}} (note: if you need help with combining fractions, check out this <a href=http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver>solver</a>)



--------------------------------


Summary:


So the equation of the tangent line is {{{y=-2x+5}}} and the equation of the normal line is {{{y=(1/2)x+5/2}}}




Notice if we graph {{{y=4-x^2}}}, the tangent line {{{y=-2x+5}}}, and the normal line {{{y=(1/2)x+5/2}}}, we get



{{{ graph( 500, 500, -10, 10, -10, 10,4-x^2, -2x+5,(1/2)x+5/2) }}} Graph of {{{y=4-x^2}}} (red), the tangent line {{{y=-2x+5}}} (green), and the normal line {{{y=(1/2)x+5/2}}} (blue)




So we can see that {{{y=-2x+5}}} is tangent to {{{y=4-x^2}}} and that {{{y=(1/2)x+5/2}}} is perpendicular to {{{y=-2x+5}}}. So this visually verifies our answer.