The normal to the curve $\displaystyle y=x^2 - 6x + 1$ at the point $\displaystyle p$ is parallel to the straight line.Find the equation of the $\displaystyle 2y = x + 3$ tangent to the curve at the point $\displaystyle p$
If this is how your post should have read, then calculating the derivative of $\displaystyle x^2-6x+1$
will give you the slope of the tangent.
The slope of the line $\displaystyle 2y=x+3\Rightarrow\ y=\frac{1}{2}x+\frac{3}{2}$
is $\displaystyle \frac{1}{2}$
The normal and the tangent are perpendicular, so you can find what the tangent slope needs to be.
The derivative being equal to this tangent slope will obtain the x co-ordinate of $\displaystyle p$
The y co-ordinate is obtained by placing the x co-ordinate into f(x).
Finally, to get the equation of the tangent, use the equation of the line formula,
using the slope calculated and the co-ordinates of $\displaystyle p$
i get:
$\displaystyle y=x^2 -3x + 5$
$\displaystyle \frac{dy}{dx} = 2x - 6$
$\displaystyle 2y = x + 3$
$\displaystyle y = \frac{1}{2}x + \frac{3}{2}$
$\displaystyle (m_1)(m_2) = -1$
$\displaystyle \frac{1}{2}(m_2) = -1$
$\displaystyle m_2 = -2$
$\displaystyle 2x - 6 = -2$
$\displaystyle x = 2$
when $\displaystyle x = 2$
$\displaystyle y = 2^2 - 6(2) + 1$
$\displaystyle y = -7$
$\displaystyle using (2,-7)$ and $\displaystyle m = -2$
$\displaystyle y + 7 = -2 (x - 2)$
$\displaystyle y + 7 = -2x + 4$
is that correct sir?
Shouldn't it say the following?
The normal to the curve $\displaystyle y=x^2 - 6x + 1$ at the point $\displaystyle p$ is parallel to the straight line $\displaystyle 2y = x + 3$. Find the equation of the tangent to the curve at the point $\displaystyle p$.
Did you copy-paste from a two-column PDF file?