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

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

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.(:

To determine the equations of two lines that pass through the point (1, -5) and are tangent to the graph of y = x^2 - 2, we need to find the slopes of these tangent lines and use point-slope form to write their equations.

Step 1: Find the derivative of the function y = x^2 - 2 to obtain the slope of the tangent line at any point on the graph.

The derivative of y = x^2 - 2 can be found using the power rule for differentiation. Take the derivative of each term separately:
dy/dx = d/dx(x^2) - d/dx(2)
= 2x - 0 (the derivative of a constant is zero)
= 2x

Step 2: Determine the slope of the tangent line at the point (1, -5) by substituting x = 1 into the derivative's equation.
m = 2(1)
m = 2

Step 3: Use the point-slope form of a linear equation to write the equation of the first tangent line.
The point-slope form is given by: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

Substituting the values into the equation:
y - (-5) = 2(x - 1)
y + 5 = 2x - 2
y = 2x - 7

So, the equation of the first line that passes through the point (1, -5) and is tangent to the graph of y = x^2 - 2 is y = 2x - 7.

Step 4: Determine the slope of the second tangent line.
Since the tangents are parallel, their slopes must be equal. Therefore, the slope of the second tangent line is also 2.

Step 5: Apply the point-slope form again to write the equation of the second tangent line.
Using the same process as before, we have:
y - (-5) = 2(x - 1)
y + 5 = 2x - 2
y = 2x - 7

So, the equation of the second line that passes through the point (1, -5) and is tangent to the graph of y = x^2 - 2 is also y = 2x - 7.

In summary, the equations of the two lines that pass through the point (1, -5) and are tangent to the graph of y = x^2 - 2 are y = 2x - 7.