Blaze Shepherd

2022-04-07

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

titemomo8gjz

Substituting x=3 in the function $f\left(x\right)=3{\left(x-2\right)}^{2}-5x+2$, we get y-coordinate of point as follows
y=f(3)
$=3{\left(3-2\right)}^{2}-5\cdot 3+2$
=-10
The slope $\frac{dy}{dx}$ of tangent to the curve: $f\left(x\right)=3{\left(x-2\right)}^{2}-5x+2$ is given by differentiating the function f(x) w.r.t. x as follows
${f}^{\prime }\left(x\right)=\frac{d}{dx}\left(3{\left(x-2\right)}^{2}-5x+2\right)$
=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 $\left({x}_{1},{y}_{1}\right)\equiv \left(3,-10\right)$ & having sloe
m=-1 is given by following formula
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
y-(-10)=-1(x-3)
x+y+7=0

maggionmoo

$f\left(x\right)=-5x+3{\left(x-2\right)}^{2}+2$ and ${x}_{0}=3$.
Find the value of the function at the given point: ${y}_{0}=f\left(3\right)=-10$
The slope of the normal line at $x={x}_{0}$ is the negative reciprocal of the derivative of the function, evaluated at $x={x}_{0}:M\left({x}_{0}\right)=-\frac{1}{{f}^{\prime }\left({x}_{0}\right)}$
Find the derivative: ${f}^{\prime }\left(x\right)={\left(-5x+3{\left(x-2\right)}^{2}+2\right)}^{\prime }=6x-17$
Hence, $M\left({x}_{0}\right)=-\frac{1}{{f}^{\prime }\left({x}_{0}\right)}=-\frac{1}{6{x}_{0}-17}$
Next, find the slope at the given point.
m=M(3)=-1
Finally, the equation of the normal line is $y-{y}_{0}=m\left(x-{x}_{0}\right)$
Plugging the found values, we get that y-(-10)=-(x-3)
Or, more simply: y=-x-7

Do you have a similar question?