Answer to Question #236239 in Calculus for luka

Question #236239

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.


1
Expert's answer
2021-09-20T06:26:21-0400
y=x24xy=x^2-4x

y=(x24x)=2x4y'=(x^2-4x)'=2x-4

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


slope1=m1=y(3)=2(3)4=2slope_1=m_1=y'(3)=2(3)-4=2

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


slope2m2=1m1=12slope_2-m_2=-\dfrac{1}{m_1}=-\dfrac{1}{2}

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


y(3)=12(x3)y-(-3)=-\dfrac{1}{2}(x-3)

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

y=12x32y=-\dfrac{1}{2}x-\dfrac{3}{2}

xx -intercept: y=0=>12x32=0=>x=3y=0=>-\dfrac{1}{2}x-\dfrac{3}{2}=0=>x=-3

A(3,0)A(-3, 0)


yy -intercept: x=0=>y=12(0)32=>y=32x=0=>y=-\dfrac{1}{2}(0)-\dfrac{3}{2}=>y=-\dfrac{3}{2}

B(0,32)B(0, -\dfrac{3}{2})



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