1. For what values of a will the straight line y=ax+1 have ONE intersection with the parabola y=-x^2-x-8?

the line intersects where

-x^2-x-8 = ax+1
x^2 + (a+1)x + 9 = 0

In order for there to be a single intersection, the discriminant must be zero. That is,

(a+1)^2 - 36 = 0

See what you can do with that.

http://www.wolframalpha.com/input/?i=plot+-x%5E2-x-8,+y%3D5x%2B1,+y%3D-7x%2B1,+-5%3C%3Dx%3C%3D5

To determine the values of "a" for which the straight line and the parabola have one intersection, we need to find the conditions under which these two functions intersect at exactly one point.

Step 1: Set the equations y = ax + 1 and y = -x^2 - x - 8 equal to each other:
ax + 1 = -x^2 - x - 8

Step 2: Simplify the equation:
x^2 + (a + 1)x + (9 + a) = 0

Step 3: For the line and the parabola to intersect at exactly one point, the quadratic equation must have only one real root. This can be achieved when the discriminant, denoted as Δ, is equal to zero. The discriminant is given by the formula: Δ = b^2 - 4ac.

In this case, the coefficients are: a = 1, b = (a + 1), and c = (9 + a).

Step 4: Calculate the discriminant and set it equal to zero:
(a + 1)^2 - 4(1)(9 + a) = 0

Step 5: Expand and simplify the equation:
a^2 + 2a + 1 - 36 - 4a = 0
a^2 - 2a - 35 = 0

Step 6: Factor the quadratic equation:
(a - 7)(a + 5) = 0

Step 7: Solve for "a" by setting each factor equal to zero:
a - 7 = 0 or a + 5 = 0

Step 8: Solve for "a":
Case 1: a - 7 = 0
a = 7

Case 2: a + 5 = 0
a = -5

Therefore, the values of "a" for which the straight line y = ax + 1 has one intersection with the parabola y = -x^2 - x - 8 are a = 7 and a = -5.

To determine the values of "a" for which the line and the parabola have one intersection, we need to find the points of intersection between the two equations.

Step 1: Set the two equations equal to each other.

ax + 1 = -x^2 - x - 8

Step 2: Rearrange the equation to place it in standard quadratic form (ax^2 + bx + c = 0).

x^2 + (a + 1)x + (9 + a) = 0

Step 3: Use the discriminant to determine the number of intersections. The discriminant (Δ) is calculated as b^2 - 4ac.

Δ = (a + 1)^2 - 4(9 + a)

For one intersection, the discriminant must be equal to zero.

Δ = 0

Step 4: Solve the equation for Δ = 0.

(a + 1)^2 - 4(9 + a) = 0

Expand and simplify:

a^2 + 2a + 1 - 36 - 4a = 0

Combine like terms:

a^2 - 2a - 35 = 0

Step 5: Solve the quadratic equation.

Using factoring or the quadratic formula, we find the values of "a" that satisfy the equation:

(a - 7)(a + 5) = 0

a - 7 = 0 or a + 5 = 0

a = 7 or a = -5

Therefore, the line y = ax + 1 will have one intersection with the parabola y = -x^2 - x - 8 for a = 7 or a = -5.