Find the equation of the tangent to y= x^2 - 6x + 3, which is perpendicular to the line x - 2y + 5 = 0
Thanks in advance
Let's put the line in a more explicit form.
$\displaystyle x - 2y + 5 = 0$
$\displaystyle \Rightarrow y = \frac {1}{2}x+ \frac {5}{2}$
The slope of the line is $\displaystyle \frac {1}{2}$, so we want the tangent line with a slope of $\displaystyle -2$, since that line is perpendicular to this one (if two lines are perpenicular, then their slopes are the negative inverses of each other).
Let's find the formula for the slope of the tangent line at any point of the curve. This is given by the derivative.
$\displaystyle y = x^2 - 6x + 3$
$\displaystyle \Rightarrow y' = 2x - 6$
This is the formula for the slope, we want the slope to be -2, so let's solve for that.
$\displaystyle 2x - 6 = -2$
$\displaystyle \Rightarrow 2x = 4$
$\displaystyle \Rightarrow x = 2$
Now when $\displaystyle x = 2$
$\displaystyle y = (2)^2 - 6(2) + 3 = -5$
So the tangent line with a slope of -2 occurs at the point (2,-5) on the curve. Now we have everything to find the equation of the tangent line.
Using (x,y) = (2,-5) and m = -2, we have by the point-slope form:
$\displaystyle y - y_1 = m(x - x_1)$
$\displaystyle \Rightarrow y + 5 = -2(x - 2)$
$\displaystyle \Rightarrow \boxed {y = -2x - 1}$