Question

The line L_(1) passes through the point P(-5,-3) and it is tangent to the graph of y=(x+3)²-7. The line L_(2) is parallel to L_(1) and it is tangent to the graph of y=5-(x-1)² Find the y-intercept of L_(2).

Answer 1

Answer: L_2=-23

Explanation:

To find the y-intercept of line L2, which is parallel to line L1 and tangent to the graph of y = 5 - (x - 1)², we need to determine the equation of line L2 first.

Since line L2 is parallel to line L1, it will have the same slope as line L1. To find the slope of line L1, we can use the derivative of the function y = (x + 3)² - 7.

Taking the derivative, we get:

dy/dx = 2(x + 3)

At the point of tangency (-5, -3), the slope of line L1 is equal to the slope of the tangent line. So we substitute x = -5 into the derivative to find the slope of line L1 at that point:

m = 2(-5 + 3) = -4

Now, we have the slope of line L2, which is -4. Using the point-slope form of a line, we can write the equation of line L2 as:

y - y1 = m(x - x1)

Substituting the point (x1, y1) = (-5, -3) and the slope m = -4, we get:

y + 3 = -4(x + 5)

Simplifying the equation, we have:

y + 3 = -4x - 20

y = -4x - 23

The y-intercept of line L2 is the value of y when x = 0. Substituting x = 0 into the equation of line L2, we find:

y = -4(0) - 23

y = -23

Therefore, the y-intercept of line L2 is -23.

I hope this helps :)