Blaze Shepherd

2022-04-07

What is the equation of the line that is normal to $f\left(x\right)=3{(x-2)}^{2}-5x+2$ at x=3?

titemomo8gjz

Beginner2022-04-08Added 10 answers

Substituting x=3 in the function $f\left(x\right)=3{(x-2)}^{2}-5x+2$ , we get y-coordinate of point as follows

y=f(3)

$=3{(3-2)}^{2}-5\cdot 3+2$

=-10

The slope$\frac{dy}{dx}$ of tangent to the curve: $f\left(x\right)=3{(x-2)}^{2}-5x+2$ is given by differentiating the function f(x) w.r.t. x as follows

${f}^{\prime}\left(x\right)=\frac{d}{dx}(3{(x-2)}^{2}-5x+2)$

=6(x-2)-5

Now, the slope of tangent at x=3,

f'(3)=6(3-2)-5

=1

hence, the slope m of normal at (3, -10) is given as

$m=-\frac{1}{{f}^{\prime}\left(3\right)}=-\frac{1}{1}=-1$

Now, equation of the normal at the point$({x}_{1},{y}_{1})\equiv (3,-10)$ & having sloe

m=-1 is given by following formula

$y-{y}_{1}=m(x-{x}_{1})$

y-(-10)=-1(x-3)

x+y+7=0

y=f(3)

=-10

The slope

=6(x-2)-5

Now, the slope of tangent at x=3,

f'(3)=6(3-2)-5

=1

hence, the slope m of normal at (3, -10) is given as

Now, equation of the normal at the point

m=-1 is given by following formula

y-(-10)=-1(x-3)

x+y+7=0

maggionmoo

Beginner2022-04-09Added 16 answers

Find the value of the function at the given point:

The slope of the normal line at

Find the derivative:

Hence,

Next, find the slope at the given point.

m=M(3)=-1

Finally, the equation of the normal line is

Plugging the found values, we get that y-(-10)=-(x-3)

Or, more simply: y=-x-7

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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