Answer to Question #236239 in Calculus for luka

Question

14. The normal to the curve y = X^2– 4x at the point ( 3 , - 3 ) cuts the X-axis at A and the

Y-axis at B. Find the equation of the normal and the coordinates of A and B.

Answer 1

$$y=x^2-4x$$

$$y'=(x^2-4x)'=2x-4$$

The slope of the tangent to the curveat the point $(3, -3)$ is

$$slope_1=m_1=y'(3)=2(3)-4=2$$

The slope of the normal to the curve at the point $(3, -3)$ is

$$slope_2-m_2=-\dfrac{1}{m_1}=-\dfrac{1}{2}$$

The equation of the normal to the curve at the point $(3, -3)$ is

$$y-(-3)=-\dfrac{1}{2}(x-3)$$

The equation of the normal to the curve at the point $(3, -3)$ in slope-intercept form is

$$y=-\dfrac{1}{2}x-\dfrac{3}{2}$$

$x$ -intercept: $y=0=>-\dfrac{1}{2}x-\dfrac{3}{2}=0=>x=-3$

$A(-3, 0)$

$y$ -intercept: $x=0=>y=-\dfrac{1}{2}(0)-\dfrac{3}{2}=>y=-\dfrac{3}{2}$

$B(0, -\dfrac{3}{2})$