calculus
posted by gt on .
Determine the equations of two lines that pass through the point (1, 5) and are tangent to the graph of y = x^2  2.

y=2x3, y=6x11

We are given a fixed point through which the tangents pass.
On the curve (parabola), the slope of the tangent is dy/dx = 2x at the point (x,x²2).
What we need here is a line that passes through a point (x,f(x)) and (1,5) with a slope of 2x.
Thus we form the equation:
(y2y1)/(x2x1)=slope
(x²2  (5))/(x1) = 2x
We solve for x=3 or x=1.
Substitute into equation
(yy0)=m(xx0)
to get the formulae alx posted. 
y = x^2  2
f' = 2x
P(1,5)
f' = slope m = 2(1) = 2
Equation of the line tangent at P(1,5)
m = 2
y = mx + b
5 = 2(1) + b
5 = 2 + b
b = 7
Equation of the tangent line is,
y = 2x  7
2x  y = 7
Equation of the normal line at P(1,5) (The normal line is the line that is perpendicular to the tangent line at the point of tangency).
m = 1/2
y = mx + b
5 = 1/2(1) + b
5 = 1/2 + b
b = 10/2 + 1/2
b = 9/2
Equation of the normal line
y = 1/2 x  9/2
2y = x  9
x + 2y = 9 
I think I read the question wrong.
I thought the question wanted the slope of the tangent of f(x) at (1,5) and the two equations at this point. 
Ignore my response.
I completely misunderstood the question.(: