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

Answer 1

y=-2x-3, y=6x-11

Answer 2

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,x²-2).

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:
(y2-y1)/(x2-x1)=slope
(x²-2 - (-5))/(x-1) = 2x
We solve for x=3 or x=-1.
Substitute into equation
(y-y0)=m(x-x0)
to get the formulae alx posted.

Answer 3

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

Answer 4

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.

Answer 5

Ignore my response.

I completely misunderstood the question.(: