Find the equation of tangent and normal for $y=f(x)$, if, $ y=x^{2}-2x+3$ and tangent is perpendicular to line $x+y-1=0$

Question

I am supposed to find the equation of tangent and normal for $y=f(x)$, if, $ y=x^{2}-2x+3$ and tangent is perpendicular to line $x+y-1=0$. My solution for tangent is $y-x-3/4=0$ and for normal is $y+x-15/4=0$. Is it correct? Thanks!

Answer 1

It's always good to check your work in math.

When you think you have an answer, go back and make sure it works to answer the question.

To check that $y-x-3/4=0$ is perpendicular to $x+y-1=0$,

put the equations in slope-intercept form $y=x+3/4$ and $y=-x+1$;

since the product of the slopes is $-1 $, these lines are perpendicular.

Likewise, the line $y=x+3/4$ is perpendicular to $y=-x+15/4.$

The latter pair of lines intersects when $x+3/4=-x+15/4$, i.e., $x=3/2.$

At that point $y=9/4.$ Furthermore, $f(3/2)=9/4$, so this line is indeed tangent to $f(x)$.