Posted by rob on Friday, July 22, 2011 at 8:27am.
dy/dx = 2x + 5
slope of tangent at (1,4) = 2(1) +5 = 7
at (1,4) ...
4 = 7(1) + b
b= -3
tangent equation at (1,4) is y = 7x - 3
at (-3,-8)
slope of tangent is 2(-3) + 5 = -1
so slope of normal is +1
equation of normal is y = x + b
at (-3,-8) , -8 = -3 + b
b = -5
equation of normal is y = x - 5
where do they intersect ?
when 7x - 3 = x - 5
6x = -2
x = -1/3
back into y = x-5 , (or the other equation, makes no differencen)
y = - 1/3 - 5 = -1/3 - 15/3 = -16/3
so the two straight lines intersect at (-1/3, -16/3)
Related Questions
Business Statistics - In each case, sketch the two specified normal ...
Maths - The tangent to the curve 2y= 2x^2 -5x +4 at the point where x=1 is ...
12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...
Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...
calculus - 1. Given the curve a. Find an expression for the slope of the curve ...
calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
calculus - A curve passes through the point (1,-11) and it's gradient at any...
calculus - Consider line segments which are tangent to a point on the right half...
math - consider the curve defined by the equation y=a(x^2)+bx+c. Take a point(h,...
Help Calc.! - original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1...
For Further Reading