1. if f(x)= 2x^2+ Ax^2 + 4x-5 and f(2)=5, what is the value of A?

2. state the domain of f(x)= x-2/x@-x-6

f(2)=5=2*2^2+A*2^2+4*2-5

5=8+4A+8-5
figure what A is.

On the second, I cant read it.

2. state the domain of f(x)= x-2/x^2-x-6

1. To find the value of A, we need to use the given equation and the fact that f(2) equals 5.

Given: f(x) = 2x^2 + Ax^2 + 4x - 5

We know that f(2) = 5, so let's substitute x = 2 into the equation:

f(2) = 2(2)^2 + A(2)^2 + 4(2) - 5

Simplifying, we have:

5 = 2(4) + A(4) + 8 - 5

Now, we can simplify further:

5 = 8 + 4A + 8 - 5

Combining like terms, we get:

5 = 4A + 11

To isolate the term with A, we subtract 11 from both sides:

5 - 11 = 4A + 11 - 11

-6 = 4A

Finally, divide both sides by 4 to solve for A:

-6/4 = 4A/4

A = -3/2 or -1.5

Therefore, the value of A is -3/2 or -1.5.

2. To determine the domain of the function f(x) = (x - 2) / (x^2 - x - 6), we need to identify any values that would result in division by zero.

The function is defined for all real numbers except those that would make the denominator equal to zero. To find these values, we set the denominator equal to zero and solve for x:

x^2 - x - 6 = 0

This quadratic equation can be factored as:

(x - 3)(x + 2) = 0

Setting each factor equal to zero gives us two possible values:

x - 3 = 0 => x = 3
x + 2 = 0 => x = -2

These are the points where the denominator becomes zero, so they are excluded from the domain.

In conclusion, the domain of the function f(x) = (x - 2) / (x^2 - x - 6) is all real numbers except x = 3 and x = -2.